\(a,ĐK:x\ne2;x\ne-3\\ b,A=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\\ A=\dfrac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\\ c,A=\dfrac{-3}{4}\Leftrightarrow4x-16=6-3x\Leftrightarrow x=\dfrac{22}{7}\left(tm\right)\\ d,A=1-\dfrac{2}{x-2}\in Z\Leftrightarrow x-2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{0;1;3;4\right\}\left(tm\right)\)
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