Quy đồng mẫu thức ba phân thức :
\(\dfrac{x}{x^2-2xy+y^2-z^2};\dfrac{y}{y^2-2yz+z^2-x^2};\dfrac{z}{z^2-2zx+x^2-y^2}\)
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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: 1/x^2y=1/x^2y
3/xy=3x/x^2y
b: \(\dfrac{x}{x^2+2xy+y^2}=\dfrac{x}{\left(x+y\right)^2}\)
\(\dfrac{2x}{x^2+xy}=\dfrac{2}{x+y}=\dfrac{2x+2y}{\left(x+y\right)^2}\)
\(\dfrac{1}{3x+xy}=\dfrac{1}{x\left(y+3\right)}=\dfrac{\left(x+y\right)^2}{x\left(y+3\right)\left(x+y\right)^2}\)
\(2x+2y=2\left(x+y\right)=\dfrac{2\left(x+y\right)\cdot x\left(y+3\right)\left(x+y\right)^2}{x\left(y+3\right)\left(x+y\right)^2}\)
\(\dfrac{1}{x^2+2xy+y^2}=\dfrac{3x+xy}{x\left(y+3\right)\left(x+y\right)^2}\)
\(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{2\cdot\left(x+y\right)}{6\left(x+y\right)^2}\)
\(\dfrac{1}{2x+2y}=\dfrac{1}{2\left(x+y\right)}=\dfrac{3\left(x+y\right)}{6\left(x+y\right)^2}\)
\(\dfrac{1}{x^2+2xy+y^2}=\dfrac{1}{\left(x+y\right)^2}=\dfrac{6}{6\left(x+y\right)^2}\)
MTC=xy(x-y)(x+2y)
\(\dfrac{x+y}{x}=\dfrac{y\left(x+y\right)\left(x-y\right)\left(x+2y\right)}{xy\left(x-y\right)\left(x+2y\right)}\)
\(\dfrac{x}{x-y}=\dfrac{x^2y\left(x+2y\right)}{xy\left(x-y\right)\left(x+2y\right)}\)
\(\dfrac{2}{x^2+2xy}=\dfrac{2}{x\left(x+2y\right)}=\dfrac{y\left(x-y\right)}{xy\left(x-y\right)\left(x+2y\right)}\)
\(\dfrac{1}{xy+2y^2}=\dfrac{1}{y\left(x+2y\right)}=\dfrac{x\left(x-y\right)}{xy\left(x-y\right)\left(x+2y\right)}\)