\(a,ĐK:x\ne-3;x\ne2\\ A=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\\ b,\left|x-3\right|=1\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\left(loại\right)\end{matrix}\right.\Leftrightarrow x=4\\ \Leftrightarrow A=\dfrac{4-4}{4-2}=0\\ c,A=1-\dfrac{2}{x-2}\in Z\\ \Leftrightarrow x-2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{0;1;3;4\right\}\left(tm\right)\)
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