viết sai đề hết rồi

rưa mi chỉnh t cấy

a: \(y^3+2xy^2+y^2-4x^2\)

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

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

\(\frac{8x^3+y^3}{y^3+2xy^2+y^2-4x^2}\)

\(=\frac{\left(2x+y\right)\left(4x^2-2xy+y^2\right)}{\left(2x+y\right)\left(y^2+y-2x\right)}=\frac{4x^2-2xy+y^2}{y^2+y-2x}\)

b: \(\frac{x^2-2x-8}{2x^2+9x+10}\)

\(=\frac{x^2-4x+2x-8}{2x^2+4x+5x+10}\)

\(=\frac{\left(x-4\right)\cdot\left(x+2\right)}{\left(x+2\right)\left(2x+5\right)}=\frac{x-4}{2x+5}\)

c: \(\frac{6x-x^2-5}{5x^6-x^7}\)

\(=\frac{x^2-6x+5}{x^7-5x^6}\)

\(=\frac{\left(x-5\right)\left(x-1\right)}{x^6\cdot\left(x-5\right)}=\frac{x-1}{x^6}\)

d: \(\frac{x^3+64}{2x^3-8x^2+32x}=\frac{\left(x+4\right)\left(x^2-4x+16\right)}{2x\left(x^2-4x+16\right)}=\frac{x+4}{2x}\)

e: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

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

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

Ta phân tích mẫu:

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

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

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

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

Thay vào ta có:

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

Vậy ta có điều phải chứng minh

tks nha <3