Đáp án: $x=3$
Giải thích các bước giải:
ĐKXĐ: $x\ge \dfrac83$
Ta có :
$2(2x-1)-3\sqrt{5x-6}=\sqrt{3x-8}$
$\to 2(2x-1)=3\sqrt{5x-6}+\sqrt{3x-8}$
$\to 2(2x-1)-10=(3\sqrt{5x-6}-9)+(\sqrt{3x-8}-1)$
$\to 2(2x-1-5)=3(\sqrt{5x-6}-3)+(\sqrt{3x-8}-1)$
$\to 2(2x-6)=3\cdot\dfrac{5x-6-3^2}{\sqrt{5x-6}+3}+\dfrac{3x-8-1}{\sqrt{3x-8}+1}$
$\to 4(x-3)=3\cdot\dfrac{5(x-3)}{\sqrt{5x-6}+3}+\dfrac{3(x-3)}{\sqrt{3x-8}+1}$
$\to x-3=0\to x=3$
Hoặc $\to 4=3\cdot\dfrac{5}{\sqrt{5x-6}+3}+\dfrac{3}{\sqrt{3x-8}+1}$
$\to 4=\dfrac{15}{\sqrt{5x-6}+3}+\dfrac{3}{\sqrt{3x-8}+1}(1)$
$+) x>3$
$\to \dfrac{15}{\sqrt{5x-6}+3}+\dfrac{3}{\sqrt{3x-8}+1}<\dfrac{15}{\sqrt{5\cdot3-6}+3}+\dfrac{3}{\sqrt{3\cdot 3-8}+1}=4$
$\to (1)$ vô nghiệm
$+) x<3\to \dfrac83\le x<3$
$\to \dfrac{15}{\sqrt{5x-6}+3}+\dfrac{3}{\sqrt{3x-8}+1}>\dfrac{15}{\sqrt{5\cdot3-6}+3}+\dfrac{3}{\sqrt{3\cdot 3-8}+1}=4$
$\to (1)$ vô nghiệm
