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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) \(\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3}=\dfrac{x^3}{\left(x-y\right)^3}=\dfrac{x^3.y}{\left(x-y\right)^3.y}=\dfrac{x^3y}{y\left(x-y\right)^3}\)
\(\dfrac{x}{y^2-xy}=\dfrac{x}{y.\left(y-x\right)}=-\dfrac{x}{y.\left(x-y\right)}=-\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right).\left(x-y\right)^2}=\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right)^3}\)
Ta có \(\frac{2}{x^3-y^3}=\frac{2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(\frac{2x-1}{x^2-y^2}=\frac{2x+1}{\left(x+y\right)\left(x-y\right)}\)
\(\frac{1}{x+y}\) giữ nguyên
MTC: \(\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
Các nhân tử phụ tương ứng là : \(\left(x+y\right);\left(x-y\right)\left(x^2+xy+y^2\right);\left(x^2+xy+y^2\right)\)
Ta có:
\(\frac{2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\frac{2.\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}\)
\(\frac{1}{x+y}=\frac{1.\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(\frac{2x+1}{\left(x+y\right)\left(x-y\right)}=\frac{\left(2x+1\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)}\)
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: 1/x^2y=1/x^2y
3/xy=3x/x^2y
b: \(\dfrac{x}{x^2+2xy+y^2}=\dfrac{x}{\left(x+y\right)^2}\)
\(\dfrac{2x}{x^2+xy}=\dfrac{2}{x+y}=\dfrac{2x+2y}{\left(x+y\right)^2}\)
Bài 1 . Chia :( x3 + 5x2 - 4x - 20) cho ( x2 + 3x - 10) ta được x+ 2
Chia :( x3 + 5x2 - 4x - 20) cho ( x2 + 7x + 10) ta được x - 2
Do đó , ta có :
\(\dfrac{1}{x^2+3x-10}=\dfrac{x+2}{\left(x^2+3x-10\right)\left(x+2\right)}=\dfrac{x+2}{x^3+5x^2-4x-20}\)
Và : \(\dfrac{x}{x^2+7x+10}=\dfrac{x\left(x-2\right)}{\left(x^2+7x+10\right)\left(x-2\right)}=\dfrac{x^2-2x}{x^3+5x^2-4x-20}\)
Bài 2 . a) Ta có :
\(\dfrac{x-1}{x^3+1}\)( giữ nguyên)
\(\dfrac{2x}{x^2-x+1}=\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{2x^2+2x}{x^3+1}\)
\(\dfrac{2}{x+1}=\dfrac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{2x^2-2x+2}{x^3+1}\)
b) Ta có MTC = x2( y - z)2
Ta có :
\(\dfrac{x+y}{x\left(y-z\right)^2}=\dfrac{x^2+xy}{x^2\left(y-z\right)^2}\)
\(\dfrac{y}{x^2\left(y-z\right)^2}\)( giữ nguyên )
\(\dfrac{z}{x^2}=\dfrac{z\left(y-z\right)^2}{x^2\left(y-z\right)^2}\)
a: \(\dfrac{1}{x+2}=\dfrac{2x-x^2}{x\left(2-x\right)\left(2+x\right)}\)
\(\dfrac{8}{2x-x^2}=\dfrac{8}{x\left(2-x\right)}=\dfrac{8x+16}{x\left(2-x\right)\left(2+x\right)}\)
b: \(x^2+1=\dfrac{x^4-1}{x^2-1}\)
\(\dfrac{x^4}{x^2-1}=\dfrac{x^4}{x^2-1}\)
c: \(\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3}=\dfrac{x^3}{\left(x-y\right)^3}=\dfrac{x^3y}{y\cdot\left(x-y\right)^3}\)
\(\dfrac{x}{y^2-xy}=\dfrac{x}{y\left(y-x\right)}=\dfrac{-x}{y\left(x-y\right)}=\dfrac{-x\left(x-y\right)^2}{y\left(x-y\right)^3}\)
Mẫu thức chung: 36 x 2 y 4
Ta có