b)\(\sqrt{3x^2-4x}=2x-3\Leftrightarrow\left(\sqrt{3x^2-4x}\right)^2=\left(2x-3\right)^2\Leftrightarrow3x^2-4x=4x^2-12x+9\Leftrightarrow3x^2-4x^2-4x+12x-9=0\Leftrightarrow-x^2+8x-9=0\Leftrightarrow-\left(x^2-8x+9\right)=0\Leftrightarrow-\left(x^2-2.x.4+16-7\right)=0\Leftrightarrow\left(x-4\right)^2=7\Leftrightarrow\left[{}\begin{matrix}x-4=-\sqrt{7}\\x-4=\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4-\sqrt{7}\\x=4+\sqrt{7}\end{matrix}\right.\left(TMĐK\right).VậyS=\left\{4-\sqrt{7};4+\sqrt{7}\right\}.\left(ĐKXĐ:3x^2-4x\ge0\Leftrightarrow x\left(3x-4\right)\ge0\Leftrightarrow x\ge\dfrac{4}{3}\right)\)
Làm nốt :))
\(a.\dfrac{\left(7-x\right)\sqrt{7-x}+\left(x-5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}=2\) ( 5 ≤ x ≤ 7 )
⇔ \(\dfrac{\left(\sqrt{7-x}+\sqrt{x-5}\right)\left[7-x-\sqrt{\left(7-x\right)\left(x-5\right)}+x-5\right]}{\sqrt{7-x}+\sqrt{x-5}}=2\)
⇔ \(\sqrt{\left(7-x\right)\left(x-5\right)}=0\)
⇔ \(x=5\left(TM\right)orx=7\left(TM\right)\)
KL........
