Đáp án: $x=1$
Giải thích các bước giải:
ĐKXĐ: $x\ge\dfrac12$
Ta có:
$2\sqrt{2x-1}+\sqrt{x+3}-\sqrt{5x+11}=0$
$\to 2(\sqrt{2x-1}-1)+(\sqrt{x+3}-2)-(\sqrt{5x+11}-4)=0$
$\to 2\cdot\dfrac{2x-1-1}{\sqrt{2x-1}+1}+\dfrac{x+3-2^2}{\sqrt{x+3}+2}-\dfrac{5x+11-4^2}{\sqrt{5x+11}+4}=0$
$\to \dfrac{4(x-1)}{\sqrt{2x-1}+1}+\dfrac{x-1}{\sqrt{x+3}+2}-\dfrac{5(x-1)}{\sqrt{5x+11}+4}=0$
$\to (x-1)(\dfrac{4}{\sqrt{2x-1}+1}+\dfrac{1}{\sqrt{x+3}+2}-\dfrac{5}{\sqrt{5x+11}+4})=0$
Lại có: $\dfrac{4}{\sqrt{2x-1}+1}+\dfrac{1}{\sqrt{x+3}+2}-\dfrac{5}{\sqrt{5x+11}+4}=0$
Mà $\sqrt{2x-1}+1<\sqrt{2x-1}+4<\sqrt{5x+11}+4$
$\sqrt{x+3}+2<\sqrt{x+3}+4<\sqrt{5x+11}+4$
$\to \dfrac{4}{\sqrt{2x-1}+1}>\dfrac{4}{\sqrt{5x+11}+4}$
$\dfrac{1}{\sqrt{x+3}+2}>\dfrac{1}{\sqrt{5x+11}+4}$
Cộng vế với vế
$\to \dfrac{4}{\sqrt{2x-1}+1}+\dfrac{1}{\sqrt{x+3}+2}>\dfrac{5}{\sqrt{5x+11}+4}$
$\to \dfrac{4}{\sqrt{2x-1}+1}+\dfrac{1}{\sqrt{x+3}+2}-\dfrac{5}{\sqrt{5x+11}+4}>0$
$\to x-1=0\to x=1$
