Đáp án: $x\in\{5,1,\dfrac{29}{9}\}$
Giải thích các bước giải:
ĐKXĐ: $\dfrac12\le x\le 5$
$11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{(5-x)(2x-1)}$
$\to 11\sqrt{5-x}+8\sqrt{2x-1}=2(5-x)+(2x-1)+15+3\sqrt{(5-x)(2x-1)}$
Đặt $\sqrt{5-x}=a, \sqrt{2x-1}=b, a,b\ge 0$
$\to 11a+8b=2a^2+b^2+15+3ab$
$\to 2a^2+b^2+15+3ab-11a-8b=0$
$\to (2a^2+2ab-6a)+(b^2+ab-3b)-(5a+5b-15)=0$
$\to 2a(a+b-3)+b(b+a-3)-5(a+b-3)=0$
$\to (2a+b-5)(a+b-3)=0$
$\to 2a+b-5=0$ hoặc $a+b-3=0$
Với $2a+b-5=0$
$\to 2\sqrt{5-x}+\sqrt{2x-1}-5=0$
$\to 2\sqrt{5-x}=5-\sqrt{2x-1}$
$\to \left(2\sqrt{5-x}\right)^2=\left(5-\sqrt{2x-1}\right)^2$
$\to 20-4x=2x+24-10\sqrt{2x-1}$
$\to -6x-4=-10\sqrt{2x-1}$
$\to \left(-6x-4\right)^2=\left(-10\sqrt{2x-1}\right)^2$
$\to 36x^2+48x+16=200x-100$
$\to 36x^2-152x+116=0$
$\to x\in\{\dfrac{29}{9},1\}$
Với $a+b-3=0$
$\to\sqrt{5-x}+\sqrt{2x-1}-3=0$
$\to \sqrt{5-x}=3-\sqrt{2x-1}$
$\to \left(\sqrt{5-x}\right)^2=\left(3-\sqrt{2x-1}\right)^2$
$\to 5-x=2x+8-6\sqrt{2x-1}$
$\to -3x-3=-6\sqrt{2x-1}$
$\to x+1=2\sqrt{2x-1}$
$\to (x+1)^2=4(2x-1)$
$\to x^2+2x+1=8x-4$
$\to x^2-6x+5=0$
$\to (x-5)(x-1)=0$
$\to x\in\{5,1\}$
