First, we use the collocation method to explain that no matter what value x takes, the value of the algebraic formula - 2x & sup2; - 4x-3 is always less than 0, and then we get the minimum value of - 2x & # 178; - 4x-3 Well, I'm wrong. It's the maximum

First, we use the collocation method to explain that no matter what value x takes, the value of the algebraic formula - 2x & sup2; - 4x-3 is always less than 0, and then we get the minimum value of - 2x & # 178; - 4x-3 Well, I'm wrong. It's the maximum

-2X & # 178; - 4x-2 + 2-3 = - 2 (X & # 178; + 2x + 1) - 1 = - 2 (x + 1) &# 178; - 1 ≤ - 1, so no matter what the real value of X is, the value of the algebraic formula - 2x & # 178; - 4x-3 is less than zero. The maximum value of the algebraic formula is - 1. Because the opening of the image parabola of the algebraic formula is downward, it has only the maximum value and no minimum value

The result of calculating the square of (Y-X) × XY / X - the square of Y is ()

The result of calculating the square of (Y-X) × XY / X - y is (- (XY) / (x + y))
(y-x)xy/(x+y)(x-y)
=-(xy)/(x+y)
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The second power of (2x + 2) - the second power factorization of X,
The quadratic factorization of the quadratic power of 1. (2x + 2) - x
Is the result (3x + 1) × (x + 1)
2.1.4 × 9-2.3 × 36
3. The fourth power of X - the second power of X
=The power of (x's power) - the power of X
=(quadratic power of X + x) × (quadratic power of X - x)
If we go on like this, how can the result be zero? I made a mistake in the first step?
4. The quadratic factorization of (5m + 2n) - the quadratic factorization of (3m + 5N)

1、 Original expansion = 4x ^ 2 + 8x + 4-x ^ 2
=3x^2+8x+4
=(x+2)(3x+2)
2、 Let's say 9, then = 9 (1.4 ^ 2 + 2.3 ^ 2 x 2 ^ 2) = 9 (1.4 ^ 2-4.6 ^ 2) = 9 x (1.4-4.6) (1.4 + 4.6) = 54 x (- 3.2) = - 172.8
3、 X ^ 4-x ^ 2 = x ^ 2 (x ^ 2-1), you are wrong in the second step
4、 You first expand the formula = 25m ^ 2 + 20Mn + 4N ^ 2-9m ^ 2-30mn-25n ^ 2 = 16m ^ 2-10mn-21n ^ 2 = (2m-3n) (8m + 7n)
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Given x = root 2 + 1, find (square of X + 1 / x-x-x / square of x-2x + 1) / 1 / X,

x=√2+1
Original formula = [(x + 1) / (X & # 178; - x) - X / (X & # 178; - 2x + 1)] / (1 / x)
={(x+1)/[x(x-1)]-x/(x-1)²}×x
={(x+1)(x-1)/[x(x-1)²]-x²/[x(x-1)²]}×x
={(x²-1-x²)/[x(x-1)²]}×x
=-1/(x-1)²
=-1/(√2+1-1)²
=-1/(√2²)
=-1/2