Đáp án: $x=1$
Giải thích các bước giải:
ĐKXĐ: $x\ge \dfrac34$
Ta có:
$\sqrt{5x+4}+\sqrt{4x-3}=\sqrt{3x+13}$
$\to (\sqrt{5x+4}+\sqrt{4x-3})^2=(\sqrt{3x+13})^2$
$\to (5x+4)+2\sqrt{5x+4}\cdot \sqrt{4x-3}+(4x-3)=3x+13$
$\to (9x+1)+2\sqrt{5x+4}\cdot \sqrt{4x-3}=3x+13$
$\to 2\sqrt{5x+4}\cdot \sqrt{4x-3}=-6x+6$
$\to \sqrt{5x+4}\cdot \sqrt{4x-3}=-3x+6$
$\to (\sqrt{5x+4}\cdot \sqrt{4x-3})^2=(-3x+6)^2$
$\to (5x+4)(4x-3)=(-3x+6)^2$
$\to 20x^2+x-12=9x^2-36x+36$
$\to 11x^2+37x-48=0$
$\to (x-1)(x+\dfrac{48}{11})=0$
$\to x-1=0$ vì $x\ge\dfrac34\to x+\dfrac{48}{11}>0$
$\to x=1$
