Two real solutions w=4 and x=5.

-1,-2,4,5

Squaring and rearranging, x²(5 +6x -x²) = 42x +40; x⁴ -6x³ -5x² +42x +40 =0 with (-20/21) ≤ x ≤ 5. Then by RRT and SDMs (x +1)(x -4)(x² -3x -10) =0; (x +1)(x -4)(x +2)(x -5) =0, that is, x = -1 (rejected) x = 4 (accepted) x = -2 (rejected) x = 5 (accepted) Thus, x = 4, 5

X>0, X=4,5

X=5,4

Got this problem

After squaring and rearranging, we get x^4-6x^3-5x^2+42x+40=0. By inspection, we see that x=5 is a solution. [x^4-6x^3-5x^2+42x+40]/(x-5) = x^3-x^2-10x-8. x^3-x^2-10x-8=0 has x=4 as a solution and [x^3-x^2-10x-8]/9x-4) = .x^2+3x+2. x^2+3x+2=0 has x=-1,-2 as solutions but by inserting them into the original equation, we see that these are spurious solutions. So, x=4,5,

Χ=4 ,χ=5

X=4,5, but I must add with disappointment that this was not at all a terrific problem

X= -1; 4; 5; -2 -1&-2 not viable Hence x= 4; 5 are only solns.

This is not real mathematics, but some kind of arranged lucky draw like cheating. I solved it very easily, as I have to believe that it has simple solutions, then I simply try and find that -1 and -2 are preliminary solutions of the quadrap equotion. Based on that, then I easily got the other two true solutions 4 and 5 as for the original equation. A new but a real challenge, we change the problem a little bit, change number 40 to 41 for instance, can anyone still solve the problem? That's why I say this is not real mathematics but arranged lucky draw, some kind of cheating game. By similar tricks, I can list hundreds of similar testing problems in one week and let thousands of mathematics genius to scratch their heads for one month!

{42x+42x ➖ }=82x^2{40x+40x ➖}=80x^2 {82x^2+80x^2}= 162x^4/{5x+5x ➖ }=10x^2 {10x+10x ➖ }=20x^2 {10x^2+20x^2}=30x^4 (x^2)^2 =x^4 {30x^4 ➖x^4}= 30x 162x^4/30x=5.12x^4 5^1.3^4x^4 1^1.3^2^2x^2^2 3^1^1x^1^2 3x^2 (x ➖ 3x+2)