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![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(M=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\dfrac{\left|x+1\right|+2x}{3x^2-2x-1}\)
a) ĐKXĐ : \(x\ne-\dfrac{1}{3};1\).
Rút gọn :
+) Khi \(x\ge-1\) :
\(M=\dfrac{x+1+2x}{\left(x+\dfrac{1}{3}\right)\cdot\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{3x+1}{\left(x+\dfrac{1}{3}\right)\cdot\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{3\cdot\left(x+\dfrac{1}{3}\right)}{\left(x+\dfrac{1}{3}\right)\cdot\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{3}{x-1}\).
+) Khi \(x< -1\) :
\(M=\dfrac{-1-x+2x}{\left(x+\dfrac{1}{3}\right)\cdot\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{x-1}{\left(x+\dfrac{1}{3}\right)\cdot\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{1}{x+\dfrac{1}{3}}\)
b) pt \(4x^2+5x-9=0\) có hai nghiệm \(\left[{}\begin{matrix}x=1\left(KTMĐK\right)\\x=-\dfrac{9}{4}\left(TMĐK\right)\end{matrix}\right.\)
Mà \(-\dfrac{9}{4}< -1\).
=> Giá trị của M : \(M=\dfrac{1}{-\dfrac{9}{4}+\dfrac{1}{3}}=-\dfrac{12}{23}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};\dfrac{1}{2};-2\right\}\)
b: \(B=\dfrac{4x^2+4x+1-4-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{2x+1}{x+2}\)
\(=\dfrac{8x-4}{2x-1}\cdot\dfrac{1}{x+2}=\dfrac{4}{x+2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)
b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)
\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)