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a: \(\dfrac{5}{2x+6}=\dfrac{5\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)}\)
3/x^2-9=6/2(x+3)(x-3)
b: \(\dfrac{2x}{x^2-8x+16}=\dfrac{2x}{\left(x-4\right)^2}=\dfrac{6x^2}{3x\left(x-4\right)^2}\)
\(\dfrac{x}{3x^2-12x}=\dfrac{x}{3x\left(x-4\right)}=\dfrac{x\left(x-4\right)}{3x\left(x-4\right)^2}\)
c: \(\dfrac{x+y}{x}=\dfrac{\left(x+y\right)\cdot\left(x-y\right)}{x\left(x-y\right)}\)
x/x-y=x^2/x(x-y)
e: \(\dfrac{1}{x+2}=\dfrac{2x-x^2}{x\left(x+2\right)\left(2-x\right)}\)
\(\dfrac{8}{2x-x^2}=\dfrac{8\left(x+2\right)}{x\left(2-x\right)\left(2+x\right)}\)
a: \(=\dfrac{27a^6b^3\cdot a^2b^6}{a^8b^8}=27b\)
b: \(=3y^2-5x^2y^3-2y^2+3x^2y^3\)
\(=y^2-2x^2y^3\)
c: \(=6x-y+2x^2+3y-2x^2+x\)
\(=7x+2y\)
d: \(=x-y+2y^2-6xy+\dfrac{10x^2}{y}\)
a: \(=\left(\dfrac{x}{y\left(x-y\right)}-\dfrac{2x-y}{x\left(x-y\right)}\right):\dfrac{x+y}{xy}\)
\(=\dfrac{x^2-2xy+y^2}{xy\left(x-y\right)}\cdot\dfrac{xy}{x+y}\)
\(=\dfrac{\left(x-y\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{x-y}{x+y}\)
b: \(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x-y\right)\left(x+y\right)}\cdot\dfrac{x-y}{2y}\)
\(=\dfrac{4xy+4y^2}{2\left(x+y\right)}\cdot\dfrac{1}{2y}=\dfrac{4y\left(x+y\right)}{4y\left(x+y\right)}=1\)
\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)
\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)
\(=\dfrac{2y+x}{y\left(2y-x\right)}+\dfrac{8x}{\left(x-2y\right)\left(x+2y\right)}+\dfrac{2y-x}{y\left(2y+x\right)}\)
\(=\dfrac{2y+x}{y\left(2y-x\right)}-\dfrac{8x}{\left(2y-x\right)\left(x+2y\right)}+\dfrac{2y-x}{y\left(2y+x\right)}\)
\(=\dfrac{\left(2y+x\right)^2}{y\left(2y-x\right)\left(2y+x\right)}-\dfrac{8xy}{y\left(2y-x\right)\left(x+2y\right)}+\dfrac{\left(2y-x\right)^2}{y\left(2y+x\right)\left(2y-x\right)}\)
\(=\dfrac{\left(2y+x\right)^2-8xy+\left(2y-x\right)^2}{y\left(2y-x\right)\left(2y+x\right)}\)
\(=\dfrac{8y^2-8xy+2x^2}{\left(y\right)\left(2y-x\right)\left(2y+x\right)}\)
Phân tích trên tử ta có:
\(=2\left(\left(2y\right)^2+4xy+x^2\right)=2\left(2y+x\right)^2\)
\(=\dfrac{2\left(2y+x\right)^2}{y\left(2y+x\right)\left(2y-x\right)}=\dfrac{2\left(2y+x\right)}{y\left(2y-x\right)}\)
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