Rút gọn \(\frac{x^2-9-\left(4x-2\right)\left(x-3\right)}{x^2-6x+9}\)
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=>\(\frac{x^2-3^2-\left(4x-2\right)\cdot\left(x-3\right)}{\left(x-3\right)^2}\)
=>\(\frac{\left(x+3\right)\cdot\left(x-3\right)-\left(4x-2\right)\cdot\left(x-3\right)}{\left(x-3\right)^2}\)
=>\(\frac{\left(x-3\right)\cdot\left(x+3-4x+2\right)}{\left(x-3\right)^2}\)
=>\(\frac{-3x+5}{x-3}\)
cho minh nhe!
1/(x^2+6x+9)-1/(x^2-6x+9)=(x-3)/(x-3)(x+3)-(x+3)/(x-3)(x+3)= -6/(x-3)(x+3)
1/(x+3)+1/(x-3)=
\(A=\left(\dfrac{-\left(x+2\right)}{x-2}+\dfrac{x-2}{x+2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{\left(2-x\right)}{\left(x-3\right)}\)
\(=\dfrac{-x^2-4x-4+x^2-4x+4-4x^2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)}{x-3}\)
\(=\dfrac{-4x^2-8x}{\left(x+2\right)}\cdot\dfrac{\left(-1\right)}{x-3}=\dfrac{4x}{x-3}\)
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ĐKXĐ \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
\(M=\frac{\left(x-3\right)^2}{2x\left(x-3\right)}\left(1-\frac{6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\)
\(=\frac{x-3}{2x}\left(1-\frac{6}{x-3}\right)\)
\(=\frac{x-3}{2x}.\frac{x-9}{x-3}=\frac{x-9}{2x}\)
\(M=\frac{\left(x-3\right)^2}{2x^2-6x}\left(1-\frac{6x+18}{x^2-9}\right)\left(x\ne\pm3;x\ne0\right)\)
\(\Leftrightarrow M=\frac{\left(x-3\right)^2}{2x\left(x-3\right)}\left(1-\frac{6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\)
\(\Leftrightarrow M=\frac{x-3}{2x}\cdot\left(1-\frac{6}{x-3}\right)\)
\(\Leftrightarrow M=\frac{x-3}{2x}\cdot\frac{x-9}{x-3}\)
\(\Leftrightarrow M=\frac{x-9}{2x}\)
Vậy với \(x\ne\pm3;x\ne0\)thì \(M=\frac{x-9}{2x}\)
\(\frac{x^2-9-\left(4x-2\right)\left(x-3\right)}{x^2-6x+9}\)
\(=\frac{\left(x-3\right)\left(x+3\right)-\left(4x-2\right)\left(x-3\right)}{\left(x-3\right)^2}\)
\(=\frac{\left(x-3\right)\left[\left(x+3\right)-\left(4x-2\right)\right]}{\left(x-3\right)^2}\)
\(=\frac{x+3-4x+2}{x-3}\)
\(=\frac{-3x+5}{x-3}\)