a/ \(\dfrac{x^2-x-6}{x^2+7x+10}=\dfrac{x^2-3x+2x-6}{x^2+2x+5x+10}=\dfrac{x\left(x-3\right)+2\left(x-3\right)}{x\left(x+2\right)+5\left(x+2\right)}=\dfrac{\left(x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x+5\right)}=\dfrac{x-3}{x+5}\)

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

a) \(\dfrac{x^2-x-6}{x^2+7x+10}=\dfrac{\left(x+2\right)\left(x-3\right)}{\left(x+2\right)\left(x+5\right)}=\dfrac{x-3}{x+5}\)

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

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

câu a đâu

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

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

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

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

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

vậy ý còn lại thì sao anh? ._.

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

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

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

1.

a)\(\left(\dfrac{1}{2}\cdot\left(-2\right)\cdot\dfrac{-1}{3}\right)\cdot\left(x^2\cdot x^2\cdot x^2\right)\cdot\left(y^2\cdot y^3\right)\cdot z\)

\(\dfrac{1}{3}x^6y^5z\)

Deg=12

Mấy câu kia tương tự nha cố gắng lên!

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

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

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

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

\(B=x-y\)\(\left(\text{Đ}K:x+2y\ne0;x+y\ne0\right)\)

Tham khảo nhé~

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

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

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

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

\(=x-y\)