\(A=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4\left(y^2-1\right)\)

   \(=\left(x-y-x-y\right)^2-4\left(y^2-1\right)\)

   \(=\left(-2y\right)^2-4y^2+4=4\)

a) \(27\left(1-x\right)\left(x^2+x+1\right)+81x\left(x-1\right)\)

\(=27\left(1-x^3\right)+81\left(x^2-x\right)\)

\(=27-27x^3+81x^2-81x\)

b) \(y\left[x^2+x\left(x-y\right)+\left(x-y\right)^2\right]+\left(x-y\right)^3\)

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

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

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

a, \(27\left(1-x\right)\left(x^2+x+1\right)+81x\left(x-1\right)=27-27x^3+81x^2-81x\)

b, \(y\left[x^2+x\left(x-y\right)+\left(x-y\right)^2\right]+\left(x-y\right)^3\)

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

a: ĐKXĐ: x<>y; x<>1/2; x<>-2

b:

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

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

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

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

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

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

Ta có: \(A=\left(\frac{x+y}{y}-\frac{2y}{y-x}\right):\frac{x^2+y^2}{y-x}+\left(\frac{x^2+1}{2x-1}-\frac{x}{2}\right)\cdot\frac{1-2x}{x+2}\)

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

Ta có: \(A=\left(x-y-1\right)^3-\left(x-y+1\right)^3+6\left(x-y\right)^2\)

\(=\left(x-y-1-x+y-1\right)\left[\left(x-y-1\right)^2+\left(x-y-1\right)\left(x-y+1\right)+\left(x-y+1\right)^2\right]+6\left(x-y\right)^2\)

\(=-2\cdot\left[3\left(x-y\right)^2+1\right]+6\left(x-y\right)^2\)

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

=-2

Bài 1:

a.\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2\left(x+y\right)\)

b.\(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=4x^2\)

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

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

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

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

\(=-xy\)

Vậy \(A=-xy\)

#\(Toru\)

a: ĐKXĐ: x<>y; x<>1/2; x<>-2

b:

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

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

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

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

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

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

Ta có: \(A=\left(\frac{x+y}{y}-\frac{2y}{y-x}\right):\frac{x^2+y^2}{y-x}+\left(\frac{x^2+1}{2x-1}-\frac{x}{2}\right)\cdot\frac{1-2x}{x+2}\)

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