b: Ta có: \(\sqrt{9x^2+12x+4}=4x\)
\(\Leftrightarrow\left|3x+2\right|=4x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x\ge-\dfrac{2}{3}\right)\\-3x-2=4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(loại\right)\end{matrix}\right.\)
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