Problem indicated is different from what is worked out
@user-bk2bi4rv4h2 сағат бұрын
問題が間違っています。 前が+、後が-
@tarunjain153710 сағат бұрын
Is he teaching maths or verses from quran?
@echandler17 сағат бұрын
While solving a quadratic for a constant is novel, it obscures the heart of the problem here. The flavor of this is that of a function and its inverse. Set a new variable y to LHS and RHS, then subtract the two resulting equations. Lastly factor and solve from there. (17+x)(17-x) = √(17^2-x) = y 17^2-x^2 = y ⇒ x^2 = 17^2- y #1 square the second set y^2 = 17^2 - x subtract and factor x^2 - y^2 = x-y (x-y)(x+y)-(x-y) = 0 (x-y)(x+y-1) = 0 Hence y=x *or* y=1-x Substitute in #1 and solve the resulting quadratics. This pattern can be applied to other similar contest problems.
@guyhoghton39919 сағат бұрын
Interesting technique, thank you. I think the following approach is simpler for this case however. Let _y = √(289 - x)_ ∴ _y² = 289 - x_ ⇒ _289 - y² = x_ ... ① The equation is: ∴ _(17 + x)(17 - x) = 289 - x² = y_ ... ② Subtract ② from ①: _x² - y² = x - y_ ⇒ _(x - y)(x + y - 1) = 0_ ⇒ _x = y or x = 1 - y_ (i) *_x = y_* ⇒ _x² = y² = 289 - x_ ⇒ _x² + x - 289 = 0_ ⇒ *_x = ½( -1 ± √[1 - (4)(-289)] ) = ½( -1 ± √1157 )_* (ii) *_x = 1 - y_* ⇒ _x = 1 - (289 - x²)_ (from ②) ⇒ _x² - x - 288 = 0_ ⇒ *_x = ½( 1 ± √[1 - (4)(-288)] ) = ½( 1 ± √1153 )_* Then filter (i) and (ii) by _-17 < x < 17_
Expanding with binomial theorem, canceling highest and lowest degree terms, and dividing through by the spurious factor of 7x (as x ≠ 0) leaves: 0 = x⁵ + 3x⁴ + 5x³ + 5x² + 3x + x. Then, noting the symmetry between paired odd and even degree terms, 0 = (x + 1) ⋅ (x⁴ + 2x³ + 3x² + 2x + 1). Now again recognizing the symmetry, and so temporarily factoring out x², gives: 0 = (x + 1) ⋅ x² ⋅ ( x² + 2x + 3 + 2(1/x) + (1/x)² ) and on regrouping 0 = (x + 1) ⋅ x² ⋅ ( [x² + (1/x)² + 2] - 2 + 2⋅[x + (1/x)] + 3 ) = (x + 1) ⋅ x² ⋅ ( [x + (1/x)]² + 2⋅[x + (1/x)] + 1 ) = (x + 1) ⋅ x² ⋅ ( [x + (1/x)] + 1 )² = (x + 1) ⋅ ( x² + x + 1 )² from which all solutions are readily obtained. One of the most important concepts you could be teaching, through demonstration, is how to present work in a clear, concise, and structured easy-to-read fashion. You tag your problems as Olympiad problems, but then side track into elementary and intermediate algebra details that the target audience is already expert in - or is working to become expert in. Help them out with that.
@key_board_x5 күн бұрын
1 + (1/x⁷) = [1 + (1/x)]⁷ 1 + (1/x)⁷ = [1 + (1/x)]⁷ → let: a = 1/x → where: x ≠ 0 1 + a⁷ = (1 + a)⁷ 1 + a⁷ = (1 + a)².(1 + a)².(1 + a)².(1 + a) 1 + a⁷ = (1 + 2a + a²).(1 + 2a + a²).(1 + 2a + a²).(1 + a) 1 + a⁷ = (1 + 2a + a² + 2a + 4a² + 2a³ + a² + 2a³ + a⁴).(1 + a + 2a + 2a² + a² + a³) 1 + a⁷ = (1 + 4a + 6a² + 4a³ + a⁴).(1 + 3a + 3a² + a³) 1 + a⁷ = 1 + 3a + 3a² + a³ + 4a + 12a² + 12a³ + 4a⁴ + 6a² + 18a³ + 18a⁴ + 6a⁵ + 4a³ + 12a⁴ + 12a⁵ + 4a⁶ + a⁴ + 3a⁵ + 3a⁶ + a⁷ 0 = 7a + 21a² + 35a³ + 35a⁴ + 21a⁵ + 7a⁶ 0 = 7.(a + 3a² + 5a³ + 5a⁴ + 3a⁵ + a⁶) a + 3a² + 5a³ + 5a⁴ + 3a⁵ + a⁶ = 0 a.(a⁵ + 3a⁴ + 5a³ + 5a² + 3a + 1) = 0 First case: a = 0 Second case: (a⁵ + 3a⁴ + 5a³ + 5a² + 3a + 1) = 0 → you can see that (- 1) is an obvious root, so you can factorize (a + 1) (a + 1).(a⁴ + αa³ + βa² + γa + 1) = 0 → you expand a⁵ + αa⁴ + βa³ + γa² + a + a⁴ + αa³ + βa² + γa + 1 = 0 → you group a⁵ + a⁴.(α + 1) + a³.(β + α) + a².(γ + β) + a.(1 + γ) + 1 = 0 → you compare to: (a⁵ + 3a⁴ + 5a³ + 5a² + 3a + 1) (α + 1) = 3 → α = 2 (β + α) = 5 → β = 3 (γ + β) = 5 → γ = 2 (1 + γ) = 3 → γ = 2 → of course, above The equation becomes: (a + 1).(a⁴ + 2a³ + 3a² + 2a + 1) = 0 (a + 1).(a⁴ + a³ + a³ + a² + a² + a² + a + a + 1) = 0 (a + 1).(a⁴ + a³ + a² + a³ + a² + a + a² + a + 1) = 0 (a + 1).[(a⁴ + a³ + a²) + (a³ + a² + a) + (a² + a + 1)] = 0 (a + 1).[a².(a² + a + 1) + a.(a² + a + 1) + (a² + a + 1)] = 0 (a + 1).(a² + a + 1).(a² + a + 1) = 0 (a + 1).(a² + a + 1)² = 0 First case: a = - 1 Second case: (a² + a + 1) = 0 Δ = (1)² - (4 * 1) = - 3 = 3i² a = - 1 ± i√3 Resume the cases: a = 0 a = - 1 a = - 1 + i√3 a = - 1 - i√3 Recall: a = 1/x → x = 1/a → where a ≠ 0 First: a = 0 → no possible because the condition Second: a = - 1 → x = 1/a = - 1 Third: a = - 1 + i√3 x = 1/(- 1 + i√3) x = (- 1 - i√3)/[(- 1 + i√3).(- 1 - i√3)] x = (- 1 - i√3)/[1 - 3i²] x = (- 1 - i√3)/4 Fourth: a = - 1 - i√3 x = 1/(- 1 - i√3) x = (- 1 + i√3)/[(- 1 - i√3).(- 1 + i√3)] x = (- 1 + i√3)/[1 - 3i²] x = (- 1 + i√3)/4 Solution = { - 1 ; (- 1 - i√3)/4 ; = (- 1 + i√3)/4 }
@pietergeerkens63245 күн бұрын
For the binomial expansion, just use Pascal's Triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 6 1 instead off all that nit-picky multiplication.
1/x = u => x = 1/u 1 + u⁷ = (1 + u)⁷ 7(u⁶ + u) + 21(u⁵ + u²) + 35(u⁴ + u³) = 0 (u⁵ + 1) + 3u(u³ + 1) + 5u²(u + 1) = 0 (u + 1)[(u⁴ - u³ + u² - u + 1) + 3u(u² - u + 1) + 5u²] = 0 (u + 1)[(u⁴ - u³ + u² - u + 1) + 3u(u² - u + 1) + 5u²] = 0 (u + 1)(u⁴ + 2u³ + 3u² + 2u + 1) = 0 u = -1 => *x = -1* u⁴ + 2u³ + 3u² + 2u + 1 = 0 u² + 1/u² + 2(u + 1/u) + 3 = 0 (u + 1/u)² + 2(u + 1/u) + 1 = 0 u + 1/u = w w² + 2w + 1 = 0 (w + 1)² = 0 w = - 1 => u + 1/u = -1 => x + 1/x = -1 x² + x + 1 = 0 *x = (-1 ± i√3)/2*
@pietergeerkens63246 күн бұрын
I believe the following is easier to follow, and with much smaller magnitude constants makes the arithmetic and algebra simpler. This is a valuable consideration under exam or contest conditions, as increasing confidence that one's algebra is correct on the first pass. Dividing through by 4 and letting u = √x / 2¹⁰ yields: ⁵√[ ½ + u ] + ⁵√[ ½ - u ] = 1 Then defining a = ⁵√[ ½ + u ] b = ⁵√[ ½ - u ] v = ab = ⁵√[ ¼ - u² ] we obtain a + b = 1 ab = v and a² + b² = (a+b)² - 2ab = 1 - 2v a³ + b³ = (a+b)³ - 3ab(a+b) = 1 - 3v and finally that 1 = a⁵ + b⁵ = (a² + b²)(a³ + b³) - (ab)²(a+b) = 1 - 5v + 6v² - v² or 5v² - 5v = 5⋅v⋅(v-1) 0. Now v = 1 can be rejected as requiring u² < 0, and v = 0 gives u² = ¼ and finally x = (2¹⁰)²⋅u² = 2¹⁸.
@SidneiMV6 күн бұрын
x + 9 = u => x = u - 9 (u - 8)³ + (u - 4)³ + u³ + (u + 4)³ + (u + 8)³ = 10³ u³ + 2(u³ + 3u8²) + 2(u³ + 3u4²) = 10³ 5u³ + 480u - 1000 = 0 u³ + 96u - 200 = 0 u³ - 8 + 96(u - 2) = 0 (u - 2)(u² + 2u + 100) = 0 u - 2 = 0 => u = 2 => *x = -7* -6³ + -2³ + 2³ + 6³ + 10³ = 10³
@pietergeerkens63246 күн бұрын
Nice problem; unimpressive solution. Always look for symmetries on complex problems! This one took only about 20 seconds to read, solve, and prove with just mental math. Here: (x+1)^3 + (x+13)^3 = 0 when x = -7 as (-6)^3 + (6)^3 = 0 (x+5)^3 + (x+ 9)^3 = 0 when x = -7 as (-2)^3 + (2)^3 = 0 Unsurprisingly, this solves the problem as 10^3 = 10^3. To succeed on Olympiad level problems, candidates must be trained to look for, and recognize, this level of analysis. The techniques you use here should be demonstrated on problems that actually require them.
Ar fi bine ca enunțul sa fie corect 160 x nu 160 Ok,
@freddyalvaradamaranon30411 күн бұрын
Interesante ejercicio de ecuacion cuadratica, gracias por compartir 😊😊. Buena y didáctica explicación, utilizando propiedades de radicales y ecuaciones cuadraticas.😊😊.
Пікірлер
Problem indicated is different from what is worked out
問題が間違っています。 前が+、後が-
Is he teaching maths or verses from quran?
While solving a quadratic for a constant is novel, it obscures the heart of the problem here. The flavor of this is that of a function and its inverse. Set a new variable y to LHS and RHS, then subtract the two resulting equations. Lastly factor and solve from there. (17+x)(17-x) = √(17^2-x) = y 17^2-x^2 = y ⇒ x^2 = 17^2- y #1 square the second set y^2 = 17^2 - x subtract and factor x^2 - y^2 = x-y (x-y)(x+y)-(x-y) = 0 (x-y)(x+y-1) = 0 Hence y=x *or* y=1-x Substitute in #1 and solve the resulting quadratics. This pattern can be applied to other similar contest problems.
Interesting technique, thank you. I think the following approach is simpler for this case however. Let _y = √(289 - x)_ ∴ _y² = 289 - x_ ⇒ _289 - y² = x_ ... ① The equation is: ∴ _(17 + x)(17 - x) = 289 - x² = y_ ... ② Subtract ② from ①: _x² - y² = x - y_ ⇒ _(x - y)(x + y - 1) = 0_ ⇒ _x = y or x = 1 - y_ (i) *_x = y_* ⇒ _x² = y² = 289 - x_ ⇒ _x² + x - 289 = 0_ ⇒ *_x = ½( -1 ± √[1 - (4)(-289)] ) = ½( -1 ± √1157 )_* (ii) *_x = 1 - y_* ⇒ _x = 1 - (289 - x²)_ (from ②) ⇒ _x² - x - 288 = 0_ ⇒ *_x = ½( 1 ± √[1 - (4)(-288)] ) = ½( 1 ± √1153 )_* Then filter (i) and (ii) by _-17 < x < 17_
{17x+17x ➖ }= 34x^2 (17x)^2=289x^2 {34x^2 ➖ 289x^2}= 255x 10^20^55 2^55^45^11 2^1^12^2^1^11^1 1^1^2^1^1 2^1 (x ➖ 2x+1) .
The trick to substitute t for 289 and solve for t is nice, but I still wouldn't call the problem "awesome".
Clever solution
Excellent substitution with new variable t. This insight isn't obvious without extensive practice.
This question demands much more math skills than knowledge
(17x)^3= 503x^3 ➖.{18x+18x+18x+18x ➖ 18x ➖ 18x}= 54x^3 {98x+98x+98x98x ➖ 98x ➖ 98x}= 294x^3 {54x^3+294x^3}=348x^6 {503x^3 ➖ 348x^6}= 165x^3 10^105^13x^3 2^52^55^13^1x^3 1^12^1^1^1^1x^3 2x^3 (x ➖ 3x+2) .
Wow. I would not have guessed that this equation had four real roots. Thank you.
2940/81+{6+6 ➖ } ={ 2940/81+12}= 2952/81= 31812 31^1 3^4 2 1^1 3^2^2 2 3^1^1 2 32. (x ➖3x+2) . 32120/8613 (6)^3= 196 {32120/8613 ➖ 196}= 31924/8613= 3912.4 100^300 10^90^12.4 10^10^10^30 10^9^10^12.4 2^5^2^52^5^5^6 2^5^92^53^4.4 1^1^1^1^1^1^1^3^2 1^13^21^13^2^2.2^2 3^2 3^23^2^2.2^2 1^1 1^13^1^1.1^2 3^1.1^2 3.2 (x ➖ 3x+2).{ 8+8 ➖}/{x^8+x^8 ➖} +{8+8 ➖}/{x^8+x^8 ➖}{ 16/x^16+16/x^16}= 32/x^32 =1 (x ➖ 1x+1) .
((3x)^2(1)^2/(x)^2) = {9x^2 ➖ 1}/x^2 =8x^2/x^2{ 3x+3x ➖ }{1+1 ➖ }/{x+x ➖ } = {6x^2+2}/x^2= 8x^2/x^2 {8x^2/x^2+8x^2/x^2 }= 16x^4/x^4 16x^1 4^4x^1^1 2^2^2^2^x 1^1^1^2x 1^2x (x ➖ 2x+1).{ 30+30 ➖ }/{x^4+3x} =60/3x^5 =20x^5 5^4x^5 1^2^2x^1 1^2 (x ➖ 2x+1).
Sorry, I meant1/x^.5 has become 2/x^,5
-1/× has become -2/× in the problem worked out.
Como hay 10 elementos en la suma pueden formarse cinco pares donde cada uno suman 47 y 5(47)=235 y se comprueba fácilmente
(x ➖ 4^3)/4x+4x+4x4x ➖ 4x ➖ 4x=12x^3+8+8+88 ➖ 8 ➖ 8=24,=64x^3/{12x^3+24}= 64x^3/36x^3 =1.28x^3 1^1.7^4x^3 7^12^2x^3^1 1^11^2x^3^1 1^2x^3^1 2x^3 (x ➖ 3x+2) .
64/{x^2+x^2 ➖ 4x^2+4x^2 ➖ }+{16+16 ➖ }= 64/{x^4+8x^4}+32= 64/{8x^8+32}= 64/40x^8 =1 .24x^8.1^1.4^6x^2^3 2^2^3^2x^2^3 1^1^3^1x^2^1 3x^2 (x ➖ 3x+2).
Expanding with binomial theorem, canceling highest and lowest degree terms, and dividing through by the spurious factor of 7x (as x ≠ 0) leaves: 0 = x⁵ + 3x⁴ + 5x³ + 5x² + 3x + x. Then, noting the symmetry between paired odd and even degree terms, 0 = (x + 1) ⋅ (x⁴ + 2x³ + 3x² + 2x + 1). Now again recognizing the symmetry, and so temporarily factoring out x², gives: 0 = (x + 1) ⋅ x² ⋅ ( x² + 2x + 3 + 2(1/x) + (1/x)² ) and on regrouping 0 = (x + 1) ⋅ x² ⋅ ( [x² + (1/x)² + 2] - 2 + 2⋅[x + (1/x)] + 3 ) = (x + 1) ⋅ x² ⋅ ( [x + (1/x)]² + 2⋅[x + (1/x)] + 1 ) = (x + 1) ⋅ x² ⋅ ( [x + (1/x)] + 1 )² = (x + 1) ⋅ ( x² + x + 1 )² from which all solutions are readily obtained. One of the most important concepts you could be teaching, through demonstration, is how to present work in a clear, concise, and structured easy-to-read fashion. You tag your problems as Olympiad problems, but then side track into elementary and intermediate algebra details that the target audience is already expert in - or is working to become expert in. Help them out with that.
1 + (1/x⁷) = [1 + (1/x)]⁷ 1 + (1/x)⁷ = [1 + (1/x)]⁷ → let: a = 1/x → where: x ≠ 0 1 + a⁷ = (1 + a)⁷ 1 + a⁷ = (1 + a)².(1 + a)².(1 + a)².(1 + a) 1 + a⁷ = (1 + 2a + a²).(1 + 2a + a²).(1 + 2a + a²).(1 + a) 1 + a⁷ = (1 + 2a + a² + 2a + 4a² + 2a³ + a² + 2a³ + a⁴).(1 + a + 2a + 2a² + a² + a³) 1 + a⁷ = (1 + 4a + 6a² + 4a³ + a⁴).(1 + 3a + 3a² + a³) 1 + a⁷ = 1 + 3a + 3a² + a³ + 4a + 12a² + 12a³ + 4a⁴ + 6a² + 18a³ + 18a⁴ + 6a⁵ + 4a³ + 12a⁴ + 12a⁵ + 4a⁶ + a⁴ + 3a⁵ + 3a⁶ + a⁷ 0 = 7a + 21a² + 35a³ + 35a⁴ + 21a⁵ + 7a⁶ 0 = 7.(a + 3a² + 5a³ + 5a⁴ + 3a⁵ + a⁶) a + 3a² + 5a³ + 5a⁴ + 3a⁵ + a⁶ = 0 a.(a⁵ + 3a⁴ + 5a³ + 5a² + 3a + 1) = 0 First case: a = 0 Second case: (a⁵ + 3a⁴ + 5a³ + 5a² + 3a + 1) = 0 → you can see that (- 1) is an obvious root, so you can factorize (a + 1) (a + 1).(a⁴ + αa³ + βa² + γa + 1) = 0 → you expand a⁵ + αa⁴ + βa³ + γa² + a + a⁴ + αa³ + βa² + γa + 1 = 0 → you group a⁵ + a⁴.(α + 1) + a³.(β + α) + a².(γ + β) + a.(1 + γ) + 1 = 0 → you compare to: (a⁵ + 3a⁴ + 5a³ + 5a² + 3a + 1) (α + 1) = 3 → α = 2 (β + α) = 5 → β = 3 (γ + β) = 5 → γ = 2 (1 + γ) = 3 → γ = 2 → of course, above The equation becomes: (a + 1).(a⁴ + 2a³ + 3a² + 2a + 1) = 0 (a + 1).(a⁴ + a³ + a³ + a² + a² + a² + a + a + 1) = 0 (a + 1).(a⁴ + a³ + a² + a³ + a² + a + a² + a + 1) = 0 (a + 1).[(a⁴ + a³ + a²) + (a³ + a² + a) + (a² + a + 1)] = 0 (a + 1).[a².(a² + a + 1) + a.(a² + a + 1) + (a² + a + 1)] = 0 (a + 1).(a² + a + 1).(a² + a + 1) = 0 (a + 1).(a² + a + 1)² = 0 First case: a = - 1 Second case: (a² + a + 1) = 0 Δ = (1)² - (4 * 1) = - 3 = 3i² a = - 1 ± i√3 Resume the cases: a = 0 a = - 1 a = - 1 + i√3 a = - 1 - i√3 Recall: a = 1/x → x = 1/a → where a ≠ 0 First: a = 0 → no possible because the condition Second: a = - 1 → x = 1/a = - 1 Third: a = - 1 + i√3 x = 1/(- 1 + i√3) x = (- 1 - i√3)/[(- 1 + i√3).(- 1 - i√3)] x = (- 1 - i√3)/[1 - 3i²] x = (- 1 - i√3)/4 Fourth: a = - 1 - i√3 x = 1/(- 1 - i√3) x = (- 1 + i√3)/[(- 1 - i√3).(- 1 + i√3)] x = (- 1 + i√3)/[1 - 3i²] x = (- 1 + i√3)/4 Solution = { - 1 ; (- 1 - i√3)/4 ; = (- 1 + i√3)/4 }
For the binomial expansion, just use Pascal's Triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 6 1 instead off all that nit-picky multiplication.
X= -1.
{1+1 ➖}{1+1 ➖}/{x^7+x^7 ➖ }={ 2+2}/x^14= 4/x^14= 2^2/x^2^7 1^1/2^7^1 21^1 2^1 (x ➖ 2x+1).{7+7}/x^7= 14/x^7 = 2b(x ➖ 2x+2) .
1/x = u => x = 1/u 1 + u⁷ = (1 + u)⁷ 7(u⁶ + u) + 21(u⁵ + u²) + 35(u⁴ + u³) = 0 (u⁵ + 1) + 3u(u³ + 1) + 5u²(u + 1) = 0 (u + 1)[(u⁴ - u³ + u² - u + 1) + 3u(u² - u + 1) + 5u²] = 0 (u + 1)[(u⁴ - u³ + u² - u + 1) + 3u(u² - u + 1) + 5u²] = 0 (u + 1)(u⁴ + 2u³ + 3u² + 2u + 1) = 0 u = -1 => *x = -1* u⁴ + 2u³ + 3u² + 2u + 1 = 0 u² + 1/u² + 2(u + 1/u) + 3 = 0 (u + 1/u)² + 2(u + 1/u) + 1 = 0 u + 1/u = w w² + 2w + 1 = 0 (w + 1)² = 0 w = - 1 => u + 1/u = -1 => x + 1/x = -1 x² + x + 1 = 0 *x = (-1 ± i√3)/2*
I believe the following is easier to follow, and with much smaller magnitude constants makes the arithmetic and algebra simpler. This is a valuable consideration under exam or contest conditions, as increasing confidence that one's algebra is correct on the first pass. Dividing through by 4 and letting u = √x / 2¹⁰ yields: ⁵√[ ½ + u ] + ⁵√[ ½ - u ] = 1 Then defining a = ⁵√[ ½ + u ] b = ⁵√[ ½ - u ] v = ab = ⁵√[ ¼ - u² ] we obtain a + b = 1 ab = v and a² + b² = (a+b)² - 2ab = 1 - 2v a³ + b³ = (a+b)³ - 3ab(a+b) = 1 - 3v and finally that 1 = a⁵ + b⁵ = (a² + b²)(a³ + b³) - (ab)²(a+b) = 1 - 5v + 6v² - v² or 5v² - 5v = 5⋅v⋅(v-1) 0. Now v = 1 can be rejected as requiring u² < 0, and v = 0 gives u² = ¼ and finally x = (2¹⁰)²⋅u² = 2¹⁸.
x + 9 = u => x = u - 9 (u - 8)³ + (u - 4)³ + u³ + (u + 4)³ + (u + 8)³ = 10³ u³ + 2(u³ + 3u8²) + 2(u³ + 3u4²) = 10³ 5u³ + 480u - 1000 = 0 u³ + 96u - 200 = 0 u³ - 8 + 96(u - 2) = 0 (u - 2)(u² + 2u + 100) = 0 u - 2 = 0 => u = 2 => *x = -7* -6³ + -2³ + 2³ + 6³ + 10³ = 10³
Nice problem; unimpressive solution. Always look for symmetries on complex problems! This one took only about 20 seconds to read, solve, and prove with just mental math. Here: (x+1)^3 + (x+13)^3 = 0 when x = -7 as (-6)^3 + (6)^3 = 0 (x+5)^3 + (x+ 9)^3 = 0 when x = -7 as (-2)^3 + (2)^3 = 0 Unsurprisingly, this solves the problem as 10^3 = 10^3. To succeed on Olympiad level problems, candidates must be trained to look for, and recognize, this level of analysis. The techniques you use here should be demonstrated on problems that actually require them.
(x ➖ 2x+1 ).
{3x^3+15x^3}= 18x^6 {27x^3+39x^3}= 63x^6 {18x^6+63x^6}=81 x^12 51x^3 {81x^12+51x^3}= 132x^15 10^10^4^8x^3^5 2^52^54^2^3x^3^5 1^1^12^21^1x3^1 2^2x^3^1 1^2x3^1 2x^3 (x ➖ 3x+2)
3√ab ≠3√a*3√b
Y=Sqrt（289-X）， Y^2=289-X， 289-X^2=Y, X^2-Y^2-(X-Y)=0
Let _y = x² + 2x - 24_ The equation is _y/x = √(3x² + 2x - 24) = √(2x² + y) (y/x ≥ 0)_ ⇒ _y/x² = √(2 + y/x²)_ Let _t = y/x²_ ∴ _t = √(2 + t)_ ⇒ _t² - t - 2 = 0_ ⇒ _(t + 1)(t - 2) = 0_ ⇒ _t = -1_ or _t = 2_ Case _t = -1:_ _y = -x²_ ⇒ _2x² + 2x - 24 = 0_ ⇒ _(x - 3)(x + 4) = 0_ _x ≠ 3_ since then _y/x = -3 < 0_ ∴ *_x = -4_* Case _t = 2:_ _y = 2x²_ ⇒ _x² - 2x + 24 = 0_ _Δ = 4 - 96 < 0_ ⇒ *_x ∉ ℝ_*
{x^5+256}= 256x^5+{x+x ➖ } =x^2 {256x^5+x^2}= 256x^7 10^20^7^8x^7 2^55^4^1^2^3x^1 1^1^12^21^1 1^2;(x ➖ 2x+1) (x^5)^2={x^25 ➖ 256 }= 231 (x)^2= x^2 {231 ➖ x^2}=229 10^20^3^2 2^55^43^2 1^1^12^23^2 1^2^3^1 23 (x ➖ 3x+2)
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Olympiad Question: x = ³√144 + ³√132 + ³√121, (x³ - 1)/x² = ? x = ³√144 + ³√132 + ³√121 = (³√12)² + (³√12)(³√11) + (³√11)² Let: a = ³√12, b = ³√11; a³ - b³ = (³√12)³ - (³√11)³ = 12 - 11 = 1 x = ³√144 + ³√132 + ³√121 = a² + ab + b² a³ - b³ = (a - b)(a² + ab + b²) = (a - b)x = 1, x = 1/(a - b) (x³ - 1)/x² = x - 1/x² = (a² + ab + b²) - (a - b)² = 3ab = 3(³√12)(³√11) = 3(³√132) = 3(5.092) = 15.275 The calculation was achieved on a smartphone with a standard calculator app
{432+396+363}》1191 11^1^91^1 1^13^30 1^1 3^5^6 1^1 3^5^3^2 1^1^3^2 1^1^ 3^1 3 1 (x ➖ 3x+1) .
i was confused by your use of the 2 for the square root. I thought at first that it was 2 times the square root .
(x^196 ➖ 196/x^8 )={ 0+0 ➖}/x^8=1/x^8 =x^8 +( x^9+9)/x^6= 9x^9/x^6=9x^1.3 {x^8+9x^1.3}= 9x^2.1 3^2.2^1 3^1.2^1 32 (x ➖ 3x+2)
(x^6+6)/(x^3+x^3)= 6x^6/x^6= 6x^1 3^2x^1^1 3^2x (x ➖ 3x+2)
(x^3)^2 ➖ (16)^2={ x^9 ➖ 256}/10 =24 7/10=204.7 10^20^4.7 2^55^4.7 2^1^12^2.7^1 1^1^11^2.1^1 21 ( x ➖ 2x+1) {480x^3+384x^3 }= 864x^6 10^808^8x^.;2^52^40 2^32^3x^6 1^11^2^201^1^1x^6 2^5^4x^6.21^2^2x^3^2 1^11^1x^3^2 x^3^2 (x ➖ 3x+2)
Ar fi bine ca enunțul sa fie corect 160 x nu 160 Ok,
Interesante ejercicio de ecuacion cuadratica, gracias por compartir 😊😊. Buena y didáctica explicación, utilizando propiedades de radicales y ecuaciones cuadraticas.😊😊.
{1+1 ➖}/{x^5+x^5 ➖ }+{1+1 ➖}/{x^10+x^10 ➖ } ={2/x^10+2/x^20}= 4/x^30 =7.2 7^1.2^1 1^1.2^1 2^1 (x ➖ 2x+1)
{x^2+x^2 ➖}= x^4 (2x)^2=4x^2 (24)^2=576 {4x^2 ➖ 576}= 572x2 {572x^2+x^4}=572x^6 10^508^9x^6 2^55^102^33^2x^6 1^1^12^5^1^1^1^1x^6 1^1x^3^2 x^3^2 (x ➖ 3x+2) {48x^2+48x^2 ➖}=96x^2 {32x+32 ➖ }= 64x^2 {96x^4+64x^2}= {160x^6 ➖ 16.056}= 15.896x^6 3^5^2^400 8^12x^6 3^5^210^40^8^12x^6,3^5^210^2^20^8^12x^6 3^5^1^10^15^4^8^12x^6 3^1^2^51^2^2^2^3^3^4x^6 11^1 1^11^1^3^2^2x^6 3^1^1x^3^2 3^1x^3^2 x^3^2 (x ➖ 3x+2)
_x³ + 2x² + 2x + 1 = 0_ ⇒ _(x³ + x²) + (x² + 2x + 1) = 0_ ⇒ _x²(x + 1) + (x + 1)² = 0_ ⇒ _(x + 1)(x² + x + 1) = 0_ ⇒ _(x + 1)(x³ - 1)/(x - 1) = 0_ since _x ≠ 1_ ⇒ _(x + 1)(x³ - 1) = 0_ ⇒ *_x = -1, ω or ω²_*
This one required math skills
I LOVE this type of Math problem! Thank You for making this video✌️!!
Very nice.
{11x+11x ➖}{ 7x+7x ➖ }/(11)^2(7)^2= {22x^2+14x^2}/{121 ➖ 49} = 36x^4/72= 2x^4; 2x^2^2 1x^1^2 x^1^2 ( x➖ 2x+1) (11)^2(7)^2/{11x+11x ➖ }{7x+7x ➖ }={121 ➖ 49}/2{2x^2+14x^2}= 72/36x^4 = 2x^4 2x^2^2 1x^1^2 x^1^2 (x ➖ 2x+1)
-⅓(1 + √19) ∊ *ℝ*
1 ,5
X(35)^2 ➖ (1)^2/(x)^2 )={ 1225 ➖ 1}/x^2 》1224/x^2 = 612 10^60^12 10^5^6^12 2^5^5^3^23^4 1^1^11^13^2^2 1^13^1^2 3^2 (x ➖ 3x+2) ((35)^2 + (2+2 ➖)/(x+x ➖)={1225+4}/x^2 = 1229/x^2 =614.1 10^60^14.1 10^5^6^14.1 10^5^3^2^14.1 2^5^53^2^2^7.1 1^1^13^1^2^1.1 32 (x ➖ 3x+2)
9^11 3^2 11^1 3^1^1^1 3^1 (x ➖ 3x+1).
{28x^4+32x^4}= 60x^8; {36x^4+40x^4} 96x^8{60x^8+96x^8}=156x^16 {156x^16+44x^4 }=200x^20 10^20x5^4 2^55^4x5^4 1^112^2x^1^12^2 1^1x2^1 x^3^1 (x ➖ 2x+1).