Given x2 + 4x-1 = 0, find 1) the 2nd power of X + the - 2nd power of X, 2) the 4th power of X + the - 4th power of X

Given x2 + 4x-1 = 0, find 1) the 2nd power of X + the - 2nd power of X, 2) the 4th power of X + the - 4th power of X

X & # 178; + 4x-1 = 0x = 0 is not the root of the equation, so the equal sign divides x left and right to get x + 4-1 / x = 0x-1 / x = - 4, the square of both sides gets X & # 178; - 2 + 1 / X & # 178; = 16x & # 178; + 1 / X & # 178; = 14 (1) the second power of X + the - second power of x = x & # 178; + 1 / X & # 178; = 14 (2) x & # 178; + 1 / X & # 178; = 14 (2) the square of both sides gets x ^ 4 +

1、 6x3-11x2 - + X + 4 can be decomposed into? X3 is the third power of X
2、 It is known that there are three integer solutions of the inequality system {x-a > 0, 1-x > 0 about X, then the value range is

Original form=
6x^3-6x^2-(5x^2-x-4)
=6x^2(x-1)-(5x+4)(x-1)
=(6x^2-5x-4)(x-1)
=(2x+1)(3x-4)(x-1)
2. From (2), X is less than 1
From (1), we get that x is greater than a, so x is 0, - 1, - 2
So the value range of a is that a is greater than - 3 and less than or equal to - 2

(1 / 2) to the 99th power x 3 to the fifth power X (- 1 / 3) to the fifth power X (- 5) to the 7th power X (- 2) to the 99th power X (1 / 5) to the 7th power

(1 / 2) to the 99th power of X3 to the fifth power of X (- 1 / 3) to the seventh power of X (- 5) to the seventh power of X (- 2) to the 99th power of X (1 / 5) = (1 / 2) ^ 99 X3 ^ 5 x (- 1 / 3) ^ 5 x (- 5) ^ 7 x (- 2) ^ 99 x (1 / 5) ^ 7 = [(1 / 2) ^ 99 x (- 2) ^ 99] x [3 ^ 5 x (- 1 / 3) ^ 5] x [(- 5) ^ 7) = [(- 1) ^ 99] x [(- 1) ^ 5]