bạn xài cái này gõ công thức ra đi

a) \(A=\left[\dfrac{x+2}{x^2-x}+\dfrac{x-2}{x^2+x}\right].\dfrac{x^2-1}{x^2-x}\)

\(A=\left[\dfrac{x+2}{x\left(x-1\right)}+\dfrac{x-2}{x\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)

\(A=\left[\dfrac{\left(x+2\right)\left(x+1\right)+\left(x-2\right)\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)

\(A=\left[\dfrac{x^2+2x+x+2+x^2-2x-x+2}{x\left(x-1\right)\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)

\(A=\dfrac{2x^2+4}{x\left(x^2-1\right)}.\dfrac{x^2-1}{x^2+2}\)

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

b) Thay \(x=-200\) vào biểu thức \(A=\dfrac{2}{x}\) ta được :

\(A=\dfrac{2}{x}=\dfrac{2}{-200}=\dfrac{-2}{200}=\dfrac{-1}{100}\)

Đkxđ : \(x+y\ne0\)

\(x^2-2y^2=xy\Rightarrow x^2-y^2=xy+y^2\)

\(\Rightarrow\left(x-y\right)\left(x+y\right)=y\left(x+y\right)\)

\(\Rightarrow x-y=y\)

\(\Rightarrow x=2y\)

Thay x = 2y vào M có :

\(M=\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)

Vậy ...

Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)

\(=\left(x+y\right)^3=1^3=1\)

Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)

Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)

\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)

\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)

\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)

\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)

b, Ta co: \(x^3+xy^2-x^2y-y^3+3\)

\(=\left(x^3-y^3\right)+\left(xy^2-x^2y\right)+3\)

\(=\left(x-y\right)^3+3xy\left(x-y\right)-xy\left(x-y\right)+3\)

= 3 ( vì x-y = 0)

1=13500

2=103500