Given that 2x-3 = 0, then the value of the algebraic formula x (square - X of x) + square (5-x) - 9 of X is

Given that 2x-3 = 0, then the value of the algebraic formula x (square - X of x) + square (5-x) - 9 of X is

2x-3=0
x=3/2
X (square of X - x) square of X (5-x) - 9
=3/2 *(9/4 -3/2) 9/4 *7/2-9
=9/8 63/8 -9
=0

If 2x-3 = 0, then the value of the algebraic formula x (square of x) + (5-x) - 9. Is

2x-3 = 0 leads to x = 1.5
X (square of x-x) + (5-x) - 9 = 1.5x (1.5 ^ 2-1.5) + (x-1.5) - 9 = -4.375

If the square of x minus 2x plus 1 equals 8, then 4x minus 8x plus 9 is equal to 8

x²-2x+1=8,
x²-2x=7
So: 4x? - 8x + 9 = 4 (x? - 2x) + 9 = 4 × 7 + 9 = 28 + 9 = 37

X + 2 / 2x + 2, the square of X / X - x-2,3 / 8-4x General fraction

The common denominator is: 4 (x + 1) (X-2)
x+2/2x+2=2(x^2-4)/4(x+1)(x-2)
The square of X / x-x-2
=4x/4(x+1)(x-2)
3/8-4x
=3(x+1)/4(x+1)(x-2)

If the square of x minus 2x = 3, find the square of 2x minus 4x + 5

Xsquare - 2x = 3
SO 2 (xsquare-2x) = 6
That is 2x square - 4x = 6
So 2x square - 4x + 5 = 11

X + 1 / 1 minus the square of X - 1 / x x + 3 times the square of X + 4x + 3 / 3 x - 2x + 1

Let me solve the problem for you. The original formula = 1 / (x + 1) - [(x + 3) / (x? - 1)] * (x-1) mm2 / (x? 4x + 3) = 1 / (x + 1) - [(x + 3) / (x? - 1)] * (x-1) 2 / [(x + 1) (x + 3)] = 1 / (x + 1) - (x-1) / (x + 1)

Given x2-x-1 = 0, find the value of the algebraic expression x3-2x + 1

∵x2-x-1=0，
∴x3-2x+1=x（x2-x-1）+（x2-x-1）+2=0+0+2=2．

If the square of X + X-1 = 0, then the value of the third power of X + the square of 2x - 7 is () a.1 b.8 C. - 6 C. - 8

x^2+x=1=x*(x+1)
x^3+x^2+x^2-7=x*x*(x+1)+x^2-7=x+x^2-7=1-7=-6

Given the square of X + X-1 = 0, find the value of the third power of the algebraic formula x + 2x-7

x^2+x-1=0
So: x ^ 2 = 1-x, and x = [(radical 5) - 1] / 2, or x = - [(radical 5) - 1] / 2
x^3+2x-7
=x(x^2+2)-7
=x(1-x+2)-7
=x(3-x)-7
=3x-x^2-7
=3x-(1-x)-7
=4x-8
=2 (Radix 5) - 10, or - 2 (Radix 5) - 10

Given x2 + X - 1 = 0, find the value of X3 + 2x2 + 3

According to the meaning of the title: x2 + x = 1,
∴x3+2x2+3，
=x3+x2+x2+3，
=x（x2+x）+x2+3，
=x+x2+3，
=4；
Or: according to the meaning of the title: x2 + x = 1,
So, X3 + 2x2 + 3,
=x3+x2+x2+3，
=x（x2+x）+x2+3，
=x+x2+3，
=1+3，
=4．