b: ĐKXĐ: \(x\ge\dfrac{3}{2}\)
Ta có: \(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{2x-3}=\sqrt{4x-4}\)
\(\Leftrightarrow2x-3=4x-4\)
\(\Leftrightarrow-2x=-1\)
hay \(x=\dfrac{1}{2}\left(loại\right)\)
d: ĐKXĐ: \(x\ge-\dfrac{3}{4}\)
Ta có: \(\dfrac{\sqrt{4x+3}}{\sqrt{x+1}}=3\)
\(\Leftrightarrow\sqrt{4x+3}=\sqrt{9x+9}\)
\(\Leftrightarrow9x+9=4x+3\)
\(\Leftrightarrow5x=-6\)
hay \(x=-\dfrac{6}{5}\left(loại\right)\)
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