$\sqrt[]{4x+1}+$ $\sqrt[]{3x-2}=5$. ĐK: $x≥-1/4$
$⇔(\sqrt[]{4x+1}+$ $\sqrt[]{3x-2})^2=5^2$
$⇔4x+1+3x-2+$$2\sqrt[]{(4x+1)(3x-2)}=25$
$⇔2\sqrt[]{12x^2-5x-2}=26-7x$ $
$⇔($ $2\sqrt[]{12x^2-5x-2})^2=(26-7x)^2$. ĐK: $x$$\leq$ $\frac{26}{7}$
$⇔4(12x^2-5x-2)=676-364x+49x^2$
$⇔48x^2-20x-8=676-364x+49x^2$
$⇔x^2-344x+684=0$=>$(x-342)(x-2)=0$
$⇔$\(\left[ \begin{array}{l}x-342=0\\x-2=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=342(loại)\\x=2(tmđk)\end{array} \right.\)
Vậy $x=2$
