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\(\left[\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right].\frac{4x^2-4}{5}\) \(ĐKXĐ:x\ne\pm1;\)
\(=\)\(\left[\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right].\frac{4\left(x^2-1\right)}{5}\)
\(=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}\right]\)\(.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\left[\frac{x^2+2x+1+6-\left(x^2+2x-3\right)}{2\left(x-1\right)\left(x+1\right)}\right].\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{10}{2\left(x-1\right)\left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5}{3}\left(x\ne y\right)\)
d) \(\dfrac{x^2-y^2}{x+y}=x-y\left(đk:x\ne-y\right)\)
e) \(\dfrac{x^3-x^2+x-1}{x^2-1}=\dfrac{x^2+1}{x+1}\left(đk:x\ne\pm1\right)\)
f) \(\dfrac{x^2+4x+4}{2x+4}=\dfrac{x+2}{2}\left(đk:x\ne-2\right)\)
a) \(\frac{1}{x^2}+y^2\)xác định\(\Leftrightarrow x^2\ne0\Leftrightarrow x\ne0\)
b)\(5x+\frac{y}{x^2}+6x+10\)
xác định\(\Leftrightarrow x^2\ne0\Leftrightarrow x\ne0\)
b