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

Sorry nhg phần này sai đề ruj Mai ak

ko nhớ cô dặn gì ak?

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

1. \(\dfrac{x^3-4x^2+4x}{x^2-4}=\dfrac{x\left(x^2-4x+4\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)}{x+2}\)

Đợi anh chút

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

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

\(b,=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\left(x\ne-y\right)\\ c,=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\left(x\ne-1;x\ne\pm2;x\ne0\right)\)

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

c: \(\dfrac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\)

b) Ta có: \(x^3-x^2y-xy^2+y^3\)

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

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

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

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

\(\left(\frac{1}{x^2-xy}-\frac{3y^2}{x^4-xy^3}-\frac{9}{x^3+x^2y+xy^2}\right).\left(y+\frac{x^2}{x+y}\right)\)

\(=\left(\frac{1}{x.\left(x-y\right)}-\frac{3y^2}{x.\left(x^3-y^3\right)}-\frac{9}{x.\left(x^2+xy+y^2\right)}\right).\left(\frac{y.\left(x+y\right)}{x+y}+\frac{x^2}{x+y}\right)\)

\(=\left(\frac{1}{x.\left(x-y\right)}-\frac{3y^2}{x.\left(x-y\right)\left(x^2+xy+y^2\right)}-\frac{9}{x.\left(x^2+xy+y^2\right)}\right).\left(\frac{y^2+xy}{x+y}+\frac{x^2}{x+y}\right)\)

\(=\left(\frac{x^2+xy+y^2}{x.\left(x-y\right)\left(x^2+xy+y^2\right)}-\frac{3y^2}{x.\left(x-y\right)\left(x^2+xy+y^2\right)}-\frac{9x-9y}{x.\left(x-y\right)\left(x^2+xy+y^2\right)}\right).\left(\frac{x^2+xy+y^2}{x+y}\right)\)

\(=\frac{x^2+xy-2y^2-9x+9y}{x.\left(x-y\right)\left(x^2+xy+y^2\right)}.\frac{x^2+xy+y^2}{x+y}\)

làm tip nha bận rồi      

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