B) Ta có: 2x-2y-x²+2xy-y²

⇔ 2(x-y)-(x²-2xy+y²)

⇔ 2(x-y)-(x-y)²

⇔ (x-y)(2-x+y)

Đúng thì tick nhé

\(a,\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)

\(=\frac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)

\(=\frac{2\left(x-2\right)}{x+2}\)

Với \(x=\frac{1}{2}\)

\(\Rightarrow\frac{2\left(x-2\right)}{x+2}=\frac{2\left(\frac{1}{2}-2\right)}{\frac{1}{2}+2}=\frac{2.-\frac{3}{2}}{\frac{5}{2}}=-3.\frac{2}{5}=\frac{-6}{5}\)

b,Do x = -5; y = 10=> y = -2x

Thay y = -2x vào biểu thức ta được

\(\frac{x^3-x^2\left(-2x\right)+x\left(-2x\right)^2}{x^3+\left(-2x\right)^3}\)

\(=\frac{x^3+2x^3+2x^2}{x^3-8x^3}\)

\(=\frac{3x^3+2x^2}{-7x^3}=\frac{3}{-7}+\frac{2}{-7x}\)

Thay x = -5 là đc

\(=\left[\left(\dfrac{-\left(x-y\right)}{x-2y}-\dfrac{x^2+y^2+y-2}{\left(x-2y\right)\left(x+y\right)}\right):\dfrac{\left(2x^2+y\right)^2-4}{x\left(x+y\right)+\left(x+y\right)}\right]:\dfrac{x+1}{2x^2+y+2}\)

\(=\dfrac{-x^2+y^2-x^2-y^2-y+2}{\left(x-2y\right)\left(x+y\right)}\cdot\dfrac{\left(x+y\right)\left(x+1\right)}{\left(2x^2+y-2\right)\left(2x^2+y+2\right)}\cdot\dfrac{2x^2+y+2}{x+1}\)

\(=\dfrac{-2x^2-y+2}{\left(x-2y\right)}\cdot\dfrac{\left(x+1\right)}{\left(2x^2+y-2\right)\left(2x^2+y+2\right)}\cdot\dfrac{2x^2+y+2}{x+1}\)

\(=\dfrac{-1}{x-2y}\)

Thay $x=-1,76$ và $y=\dfrac{3}{25}$ vào $P=\dfrac{-1}{x-2y}$, ta được:

$P=\dfrac{-1}{-1,76-2.(\dfrac{3}{25})}=\dfrac{1}{2}$.

a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)

b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)

c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)

d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)

e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)