MT1: x – y

MT2: 1

MTC: x – y

NTP1: 1;    NTP2: x – y.

Quy đồng: 

a: 1/x^2y=1/x^2y

3/xy=3x/x^2y

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

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

Mik cảm ơn ạ

a: \(\frac{x+y}{x^2\left(y+z\right)}=\frac{\left(x+y\right)\cdot y^2z^2\left(x+z\right)\left(x+y\right)}{x^2y^2z^2\left(x+y\right)\left(y+z\right)\left(x+z\right)}=\frac{\left(x+z\right)\cdot y^2z^2\left(x+y\right)^2}{x^2y^2z^2\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)

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

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

b: \(\frac{5x}{x^2+5x+6}=\frac{5x}{\left(x+2\right)\left(x+3\right)}=\frac{5x\left(x+5\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)

\(\frac{2x+3}{x^2+7x+10}=\frac{2x+3}{\left(x+2\right)\left(x+5\right)}=\frac{\left(2x+3\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)

\(-5=\frac{-5\left(x+2\right)\left(x+3\right)\left(x+5\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)

Bài giải

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

\(\dfrac{8}{2x-x^2}=\dfrac{8}{x\left(2-x\right)}=-\dfrac{8}{x\left(x-2\right)}=-\dfrac{8.\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)

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

\(\dfrac{x^4}{x^2-1}\) giữ nguyên.

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

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

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

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

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

\(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{2\cdot\left(x+y\right)}{6\left(x+y\right)^2}\)

\(\dfrac{1}{2x+2y}=\dfrac{1}{2\left(x+y\right)}=\dfrac{3\left(x+y\right)}{6\left(x+y\right)^2}\)

\(\dfrac{1}{x^2+2xy+y^2}=\dfrac{1}{\left(x+y\right)^2}=\dfrac{6}{6\left(x+y\right)^2}\)

đề viết là 1/2y+2x mà bạn