giúp em nhanh với ạ. toán 9 ạ
\(A.\left(\sqrt{x}+6\right)=B.\left|x-36\right|\left(x\ge0;x\ne4\right)\)
\(\Leftrightarrow\dfrac{x-6}{\sqrt{x}-2}.\left(\sqrt{x}+6\right)=\dfrac{-1}{\sqrt{x-2}}.\left|x-36\right|\)
\(\Leftrightarrow\left(x-6\right)\left(\sqrt{x}+6\right)=-\left|x-36\right|\left(1\right)\)
\(TH_1:x-36\ge0\Leftrightarrow x\ge36\)
\(\left(1\right)\Leftrightarrow\left(x-6\right)\left(\sqrt{x}+6\right)=-\left(x-36\right)\)
\(\Leftrightarrow\left(\sqrt{x}+6\right)\left(x-6+\sqrt{x}-6\right)=0\)
\(\Leftrightarrow x+\sqrt{x}-12=0\left(\sqrt{x}+6>0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-4\left(loại\right)\\\sqrt{x}=3\left(loại\right)\end{matrix}\right.\) \(\Leftrightarrow PTVN\)
\(TH_2:x-36< 0\Leftrightarrow x< 36\)
\(\left(1\right)\Leftrightarrow\left(x-6\right)\left(\sqrt{x}+6\right)=x-36\)
\(\Leftrightarrow\left(\sqrt{x}+6\right)\left(x-6-\sqrt{x}+6\right)=0\)
\(\Leftrightarrow x-\sqrt{x}=0\left(\sqrt{x}+6>0\right)\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\) thỏa mãn \(x< 36\)
Vậy \(x\in\left\{0;1\right\}\) thỏa mãn yêu cầu đề bài
