When x = 1 is, the value of the square - 2x + a of the algebraic expression x is 3. When = - 1, the value of the square - 2x + a of the algebraic expression x is obtained

When x = 1 is, the value of the square - 2x + a of the algebraic expression x is 3. When = - 1, the value of the square - 2x + a of the algebraic expression x is obtained

Just put it in
When x = 1, then 1-2 + a = 3, so a = 4
So when x = - 1, the formula is 1 + 2 + 4 = 7

Given the square of X - 2x = 3, find the value of the square + (x + 2) (X-2) + (x-3) (x-1) of the algebraic expression (x-1)

x²-2x=3
(x-1)²+(x+2)(x-2)+(x-3)(x-1)
=x²-2x+1+x²-4+x²-4x+3
=3x²-6x+4
=3(x²-2x)+4
=9+4
=13

If the value of the algebraic expression X2 is equal to the value of the algebraic expression 2x + Y-1, then the value of the algebraic expression 9-2 (y + 2x) + 2x2 is () A. 7 B. 4 C. 1 D. Not sure

∵ the value of the algebraic expression X2 is equal to that of the algebraic expression 2x + Y-1,
∴x2=2x+y-1；
∴x2-（2x+y）=-1；
∴2y=2x2-4x+2
∴9-2（y+2x）+2x2=9-2y-4x+2x2=9-（2x2-4x+2）-4x+2x2=9-2x2+4x-2+2x2-4x=9-2=7．
Therefore, a

Given x + 2 = 1 / x, try to find the value of the algebraic formula 1 / (x + 1) - (x + 3) / (square of X - 1) × (square of X - 2x + 1) / (square of X + 4x + 3) Fractional equation

X + 2 = 1 / XX? + 2x = 11 / (x + 1) - (x + 3) / (square of x-1) × (square of x-2x + 1) / (square of X + 4x + 3) = 1 / (x + 1) - (x + 3) / (x + 1) (x-1) x (x-1) / (x + 1) (x + 3) = 1 / (x + 1) - (x-1) / (x + 1) Ω = (x + 1-x + 1) / (x + 1) Ω = 2 / (x + 1) mm2 = 2 / (x + 1) Ω = 2 / (x + 1) ·

Given x = 2-radical 3, find the value of (x-1 / 1-2x + x-squared) - (x-square-x-cent root x-square-2x + 1)

The original formula = (1-2x + X / / (x-1) / (x-1) / (x-1) / (x-1) / (x-1) / (x-1) / (x-1) / (x-1) / (x-1) / (x-1) / (x-1) / [x (x-1)] = X-1 - | X-1 (x-1) / [x (x-1)] = X-1 + (x-1) / [x (x-1)] = X-1 + (x-1)] = X-1 + (x-1) / [x (x-1)] = X-1 + (1 / x = 1 / x = 1 / x = 1-3 + 1 / (2-3) = 1 / 3) = 1-3 + 1 / (2 - / 3) = 3 1 - √ 3 + (2 + √ 3) / [(2 + √ 3

The square root of X + X + 1

[x squared - x parts x + 1-x square - 2x + 1 / x] divided by 1 / X
=[(x+1)/x(x-1)-x/(x-1)²]×x
=(x+1)/(x-1)-x²/(x-1)²
=(x²-1-x²)/(x-1)²
=-1/(x-1)²
=-1/2;
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Given that y = 2x-3 + 3-2x + 5, find the square root of X + y + 5

∵2x-3≥0 3-2x≥0
∴x=3/2
∴y=5
∴±√(x+y+5/2)=±√(3/2+5+5/2)=±√9=±3

Given x = - 2, find (1-1 / x) divided by the square of x-2x + 1

(1-1 / x) divided by the square of X / X - 2x + 1
=(x-1) x (x-1) 2 / X
=(x-1) 1 / 2
=(- 2-1) 1 / 2
=-1 / 3

Given the square of X + 3x-1 = 0, find the value of square + (1 / 2 of x) + 2x - (2 / x)

3x-1 of the square
x-1/x=-3
Square of X + (1 / 2 of x) + 2x - (2 / x)
(x-1/x)^2+2(x-1/x)+2
=(-3)^2-2*-3+2
=17

Given that x = - √ 2, find the value of (x's Square - 1) part (x's Square - 2x + 1) / (x + 1) part (x's Square - x) - X's 1 + X's square value

(x squared-1) divided (x squared - 2x + 1) / (x + 1) divided (x squared - x) - x divided into 1 + X squared
=(x²-2x+1)/(x²-1)÷(x²-x)/(x+1)-(1+x²)/x
=(x-1)²/(x+1)(x-1)÷x(x-1)/(x+1)-(1+x²)/x
=(x-1)/(x+1)×(x+1)/x(x-1)-(1+x²)/x
=1/x-(1+x²)/x
=(1-1-x²)/x
=-x
Because x = - √ 2
So the original formula = √ 2