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

3/xy=3x/x^2y

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

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

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

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

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

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

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

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

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

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

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

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

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

a) \(\dfrac{1}{x^3-8}=\dfrac{1}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{2}{2\left(x-2\right)\left(x^2+2x+4\right)}\)

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

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

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

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

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

\(\dfrac{5}{2-x}=\dfrac{-5}{x-2}=\dfrac{-5\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)^2}\)

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

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

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

Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

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

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

d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)

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

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

*Phân tích thành nhân tử:

x+2

2x-x²=x(2-x)

*MTC:x(2-x)(x+2)

*NTP:x(2-x);x+2

*Quy đồng

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

\(\dfrac{8}{x\left(2-x\right)}\)\(\dfrac{8\left(x+2\right)}{x\left(2-x\right)\left(x+2\right)}\)=\(\dfrac{8x+16}{x\left(2-x\right)\left(x+2\right)}\)

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

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

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

b) \(x^3-x+3x^2y+3xy^2+y^3-y\)

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

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

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

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

nhóm 1: \(\dfrac{5}{3}x^2y;\dfrac{-1}{3}x^2y;x^2y;\dfrac{-2}{5}x^2y\)

nhóm 2: \(xy^2;-2xy^2;\dfrac{1}{4}xy^2\)

nhóm 3: xy

2.\(25xy^2+55xy^2+75xy^2=155xy^2\)

3. thay x=1 và y=-1 vào biểu thức ta đc:

\(\dfrac{1}{2}.1^5.\left(-1\right)-\dfrac{3}{4}1^5.\left(-1\right)1^5.\left(-1\right)=\dfrac{1}{4}\)