Bài 5:
ĐKXĐ: \(x\notin\left\{-\dfrac{2}{3};3\right\}\)
Ta có: \(\dfrac{6x-1}{3x+2}=\dfrac{2x+5}{x-3}\)
\(\Leftrightarrow\left(6x-1\right)\left(x-3\right)=\left(3x+2\right)\left(2x+5\right)\)
\(\Leftrightarrow6x^2-18x-x+3=6x^2+15x+4x+10\)
\(\Leftrightarrow-19x+3=19x+10\)
\(\Leftrightarrow-19x+3-19x-10=0\)
\(\Leftrightarrow-38x-7=0\)
\(\Leftrightarrow-38x=7\)
hay \(x=-\dfrac{7}{38}\)
Vậy: \(x=-\dfrac{7}{38}\)
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