Câu hỏi của Ame No Hi

Question

Giúp mình câu c với ạ, mình cần gấp

Nguyễn Hoàng Minh · Accepted Answer

$a,A=\dfrac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x^2+9x}{\left(x-3\right)\left(x+3\right)}\
A=\dfrac{-3x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x}{x+3}\
b,\left|x-2\right|=1\Leftrightarrow\left[{}\begin{matrix}x-2=1\2-x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\x=1\left(nhận\right)\end{matrix}\right.\
\Leftrightarrow A=\dfrac{-3\cdot1}{1+3}=\dfrac{-3}{4}\
c,A=\dfrac{-3\left(x+3\right)+9}{x+3}=-3+\dfrac{9}{x+3}\in Z\
\Leftrightarrow x+3\inƯ\left(9\right)=\left\{-9;-3;-1;1;3;9\right\}\
\Leftrightarrow x\in\left\{-12;-6;-4;-2;0;6\right\}\left(tm\right)$Câu trả lời: $a,A=\dfrac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x^2+9x}{\left(x-3\right)\left(x+3\right)}\
A=\dfrac{-3x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x}{x+3}\
b,\left|x-2\right|=1\Leftrightarrow\left[{}\begin{matrix}x-2=1\2-x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\x=1\left(nhận\right)\end{matrix}\right.\
\Leftrightarrow A=\dfrac{-3\cdot1}{1+3}=\dfrac{-3}{4}\
c,A=\dfrac{-3\left(x+3\right)+9}{x+3}=-3+\dfrac{9}{x+3}\in Z\
\Leftrightarrow x+3\inƯ\left(9\right)=\left\{-9;-3;-1;1;3;9\right\}\
\Leftrightarrow x\in\left\{-12;-6;-4;-2;0;6\right\}\left(tm\right)$

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