Bài 1:

a) \(\dfrac{5x-6}{2x-1}=\dfrac{3x}{x-2}\)      (1)

ĐK:\(\left\{{}\begin{matrix}2x-1\ne0\\x-2\ne0\end{matrix}\right.\) ⇔\(\left\{{}\begin{matrix}x\ne\dfrac{1}{2}\\x\ne2\end{matrix}\right.\)

(1) ⇒\(\dfrac{\left(5x-6\right)\left(x-2\right)}{\left(2x-1\right)\left(x-2\right)}=\dfrac{3x\left(2x-1\right)}{\left(2x-1\right)\left(x-2\right)}\)

⇔\(5x^2-10x-6x+12=6x^2-3x\)

⇔\(5x^2-10x-6x+12-6x^2+3x=0\)

⇔\(-x^2-13+12=0\)

⇔\(\left[{}\begin{matrix}x=\dfrac{-13+\sqrt{217}}{2}\\x=\dfrac{-13-\sqrt{217}}{2}\end{matrix}\right.\left(TM\right)\)

KL:

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Bài 1: b) \(\dfrac{7x}{2x+2}=\dfrac{3x+1}{x-5}\)     (2)

ĐK: \(\left\{{}\begin{matrix}2x+2\ne0\\x-5\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne5\end{matrix}\right.\)

(2)⇒ \(\dfrac{7x\left(x-5\right)}{\left(2x+2\right)\left(x-5\right)}=\dfrac{\left(3x+1\right)\left(2x+2\right)}{\left(2x+2\right)\left(x-5\right)}\)

\(\Leftrightarrow7x^2-35x=6x^2+6x+2x+2\)

\(\Leftrightarrow7x^2-35x-6x^2-6x-2x-2=0\)

\(\Leftrightarrow x^2+27x-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-27+\sqrt{737}}{2}\\x=\dfrac{-27+\sqrt{737}}{2}\end{matrix}\right.\left(TM\right)\)

KL: