\(a,ĐK:x\ne\pm3\\ b,P=\dfrac{x^2-4x+3+2x+6-x^2-3}{\left(x-3\right)\left(x+3\right)}\\ P=\dfrac{-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2}{x+3}\\ c,\left|x-2\right|=1\Leftrightarrow\left[{}\begin{matrix}x=1+2=3\left(ktm\right)\\x=-1+2=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\\ \Leftrightarrow P=\dfrac{-2}{1+3}=-\dfrac{1}{2}\\ c,P\in Z\Leftrightarrow x+3\inƯ\left(-2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{-5;-4;-2;-1\right\}\)
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