Another approach let a = x + 8.5, b = x - 8.5, (a - 3/2)*(a - 1/2)*(a + 1/2)*(a + 3/2) = (b + 3/2)*(b + 1/2)*(b - 1/2)*(b - 3/2) (a^2 - 1/4)*(a^2 -9/4) = (b^2 - 1/4)*(b^2 -9/4) a^4 - b^4 - 10/4*(a^2 - b^2) = (a^2 + b^2)*(a^2 -b^2) - 10/4*(a^2 - b^2) = 0 (a^2 -b^2)*(a^2 + b^2-10/4) = (a - b)*(a + b)*(a^2 + b^2 - 10/4) = 0 (case a - b = 0) since a≠b, no solution (case a + b = 0) a + b = 2x = 0 => x = 0 (case a^2 + b^2 - 10/4 = 0) a^2 + b^2 - 10/4 = 2*x^2 + 568/4 = 0, no real solutions x = 0 is the only real solution.

No need of ∆, X^2 = -71 so X is no real number (X = i√71)

(x+7)(x+8)(x+9)(x+10)=(x-7)(x-8)(x-9)(x-10) x^4+Ax^3+Bx^2+Cx+D=x^4-Ax^3+Bx^2-Cx+D A,B,C and D =constant ; Positive integer 2Ax^3+2Cx=0 2x(Ax^2+C)=0 ⇒x=0

three cases x>0, x=0 and x0, |fraction|>1 and if x

@paseptquatre3137

2 ай бұрын

The smartest way to solve it, according to me

In the verification you dont need to multiply and since in the dominator there are 4 minus so the dominator will be positive

It's essentially a symmetrical problem when you move the entire denominator to the right side. You can see by inspection that the answer is X = 0. For any (X + n) = (X - n) when X = 0, n = -n .

(x+7)(x+8)(x+9)(x+10)=(x-7)(x-8)(x-9)(x-10), 68x^3+4828x=0, x^3+71x=0, x(x+i sqrt(71))(x-i sqrt(71))=0, x=0 or x=i sqrt(71) or x=-i sqrt(71).

Числитель и знаменатель равны между собой, без этого условия 1 не выйдет при делении. Соответственно решается уже не дробь, а равенство

at 11:25 we have (8/-8) = -1; (9/-9)=-1. etc. Four (-1)s. Their product is 1.

Я просто раскрыл все скобки. Получилось, что X^4 + 34X^3 + 341X^2 + 2414X + 5040 = X^4 - 34X^3 + 341X^2 - 2414X + 5040 X^4, X^2 и свободный член сократились. Осталось уравнение: 68X^3 + 4828X = 0. Отсюда X1 = 0 Сокращаем X и получаем, что X^2 = -71. X2 = (√71)i Но так как мы рассматривали множество только рациональных чисел, то X2 не удовлетворяет условию.

Use wavy curv

Observe that the nominator and denominator • consists of 4 factors • are identical, except the sign: + for nominator but - for denominator Therefore for x=0 the nominator and denominator are the same: 7(8)(9)(10)/[(-7)(-8)(-9)(-10)] Thus x=0 is the solution.

Also, the limit of the expression: [(X+7)(X+8)(X+9)(X+10) / (X-7)(X-8)(X-9)(X-10)] as X tends to infinity, is 1.

Why did I watch 12 minutes long overcomplicated solution for problem, which I solved without calculator or paper for about 15 seconds?

@gordongekko9742

2 ай бұрын

How?

@NESqVic

2 ай бұрын

@@gordongekko9742 Presented solution is way too overcomplicated. Ask your math teacher if he can find simpler solution.

@BlaqRaq

2 ай бұрын

You were probably hoping to see if there were any other solution he might find. Sorry. I feel your pain.

@NESqVic

2 ай бұрын

@@BlaqRaq Exactly.

@LS-Moto

2 ай бұрын

​@@gordongekko9742A fraction is only equal to 1, when both the numerator and denominator are the same. So you can just go ahead and set numerator = denominator and then solve for x

This is olympiad? Even powers of both polynomials cancel out do you are left with third degree equation without constant. Divide by x and you get quadratic. You can solve quadratic if you attend olympiad I assume :)

I solved it in 7 seconds

You could have cross multiplied much earlier

At a certain point, I feared "i". Shivers down my spine! Fortunately, in this case x€R... thanx

I will share with my students but will not restrict x to real numbers.

@NikitaP74351

2 ай бұрын

Тогда будет решение, что X = (√71)i. Нашел через раскрытие скобок, калькулятор комплексных чисел действительно говорит, что будет 1

first, x can not be 7,8,9,or10. then that x=0 is clear without putting too much effort.

(x+7) (x + 10) /( x -7) ( x -10) = ( x - 8) ( x - 9) / (x + 8) ( x + 9) (x^2 +17 x + 70) /(x^2 -17 x + 70) = (x^2 -17 x + 72) / (x^2 -17 x + 72) (x^2 + 70) /(17 x) = (x^2 + 72) /(-17 x) if x NOT EQUAL TO ZERO x^2 + 71 = 0. ELSE x = 0 ONLY REAL SOLUTION IS x =0

👍👍👍

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It’s easy problem: to get 1, we need the numerator and denominator be equal in values and signs and the denominator can’t t be equal to 0, so x=7 ,x=8,x=9,x=10 have to illuminated.; so x+7=x-7; x+8=x-8; x+9=x-9;x+10=x-10; any value for x doesn’t work,except x=0, but =7 do not equal -7, unless signs for numerator and denominator are the same; if x=0 , then (-7)*(-8)*(-9)*(-10) give us sign +, because of even number of negative as multiples,and numerator is already for x=0 is positive. So x=0 is the answer for this particular equation.

@bjornfeuerbacher5514

2 ай бұрын

"so x+7=x-7; x+8=x-8; x+9=x-9;x+10=x-10" That does not follow, why do you think so?

@larisamedovaya9097

2 ай бұрын

@@bjornfeuerbacher5514 I just picked up the pairs, but you want the hard way,just multiply every expression in the parenthesis, then make equal variables with the same power and solve them as the system of equations. You can do in different way by multiplying in pairs (…)*(…) and make quadratic expression at the left of equality to the expression at the right side and solve by solving quadratic equation,where on the left side will be quadratic equation and after equality sign “0”. I also check my answer to be true solution. Moreover, the Rule: if number of multiples is even, result will be always positive (for negative sign); example (-1*(-3)*(-4)- number of negatives is 3- odd, result will be (-), -(1*3*4)=-12; but (-1)*(-3)*(-1)*(-4)= +12, because number of negative signs is even, so you don’t multiply numbers in pairs to get final sign. I hope I have explained my solution way thoroughly.

@larisamedovaya9097

2 ай бұрын

@@bjornfeuerbacher5514 Let use example: x^2*(x+1)=12; we can break 12=4*3 and x^2=4 and X+1=3; solution x=2 and check; x^2=3 and x+1=4 doesn’t work or you can multiply x^2*(x+1)-12= and get x^3+x^2-12=0 and solve by checking x=1; x=-1; x=-2; x=2..And since cubic equation has three solutions divide x^3+x^2-12 by (x-2) .

@larisamedovaya9097

2 ай бұрын

@@bjornfeuerbacher5514 When you are looking for solution, it’s not necessary follow the standard way to do work, maybe some tricks give you easy but not standard solution. Example: Integral of(x^2)/(x+1)dx- standard way is long( decomposion), but add to numerator (+1) +(-1), we can brake our integral expression by Integral (x^2-1)/(x+1) and Integral of +1/(x+1)dx ,and using restriction x not equal -1( denominator can’t be equal 0, we simplify to Integral of (x-1)=Integral of (x)dx -Integral of dx; now to everything we can apply standard formuli. The same we can do to 1 for multiplication because for example,1=(5)*(1/5) is the same es 5/5

(x+7) (x+8) (x+9) (x+10) = (x-7) (x-8) (x-9) (x-10) only if (x+7) (x+8) (x+9) (x+10) = |(x-7)| |(x-8)| |(x-9)| |(x-10)| (x+7) (x+8) (x+9) (x+10) = [-(x-7)] [-(x-8)] [-(x-9)] [-(x-10)] (x+7) (x+8) (x+9) (x+10) = (-x+7) (-x+8) (-x+9) (-x+10) x = -x only if x =0

@bjornfeuerbacher5514

2 ай бұрын

From (x+7) (x+8) (x+9) (x+10) = (-x+7) (-x+8) (-x+9) (-x+10), it does _not_ follow directly that x has to be equal to -x.

@lechaiku

2 ай бұрын

​@@bjornfeuerbacher5514 the RHS has also 4 factors and what is more important the each sum (as a factor) has the same addends with the same uknown (x) as the LHS that means you can not change the orders of factors. The only solution is: x +7 = -x + 7 2x = 7-7 x = 0 and the same with x +8 = -x + 8 and x +9 = -x + 9 and x +10 = -x + 10

x=0.

Rearrange as: (x+7)(x+9) (x-8)(x-10) ---------------- = --------------------- (x-7)(x-9) (x+8)(x+10) (x+8 -1)(x+8 +1) (x-9 +1)(x-9 -1) -------------------------- = ----------------------------- (x-8 +1)(x-8 -1) (x+9 -1)(x+9 +1) (x+8)^2-1 (x-9)^2 -1 --------------- = ------------------ (x-8)^2-1 (x+9)^2 -1 Since (a+b)^2 = (a-b)^2 + 4(a b) we get (x-8)^2-1 + 4(8 x) (x+9)^2 -1 - 4(9 x) --------------------------- = ----------------------------- (x-8)^2-1 (x+9)^2 -1 32 x - 36 x 1 + ----------------- = 1 + ------------------ (x-8)^2-1 (x+9)^2 -1 32 x ( (x+9)^2 -1 ) = - 36 x ( (x-8)^2 -1 ) 4 x (8 (x+9)^2 - 8 + 9 (x-8)^2 - 9) = 0 x( 8 (x^2 + 18 x + 81) -8 + 9 (x^2 -16 x + 64) - 9) = 0 x( 17 x^2 + (8*18 - 9*16) x + 8 * 81 + 9 * 64 - 17) = 0 x( 17 x^2 + 1207 ) = 0 x = 0 -------> [Answer] Or 17 x^2 + 1207 = 0 x^2 = -71 x = +/- i √71 Rejected

X= 0

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