x^2 - 4x + 1 - ∛3x + 5 = √3x + 1

Giải thích các bước giải:

Ta có :
$x^2-4x+1-\sqrt[3]{3x+5}=\sqrt{3x+1}$

Đặt $\sqrt[3]{3x+5}=u\to x=\dfrac{u^3-5}{3}$

$\to \left(\dfrac{u^3-5}{3}\right)^2-4\left(\dfrac{u^3-5}{3}\right)+1-u=\sqrt{3\left(\dfrac{u^3-5}{3}\right)+1}$
$\to \dfrac{u^6}{9}-\dfrac{22u^3}{9}-u+\dfrac{85}{9}+1=\sqrt{3\cdot \dfrac{u^3-5}{3}+1}$

$\to \left(\dfrac{u^6}{9}-\dfrac{22u^3}{9}-u+\dfrac{85}{9}+1\right)^2=\left(\sqrt{3\cdot \dfrac{u^3-5}{3}+1}\right)^2$

$\to \dfrac{u^{12}}{81}-\dfrac{44u^9}{81}-\dfrac{2u^7}{9}+\dfrac{224u^6}{27}+\dfrac{44u^4}{9}-\dfrac{4136u^3}{81}+u^2-\dfrac{188u}{9}+\dfrac{8755}{81}+1=u^3-4$

$\to u\approx \:2.73156\dots ,\:u\approx \:1.61705\dots$

$\to \sqrt[3]{3x+5}=2.73156\dots$ hoặc $\sqrt[3]{3x+5}=1.61705\dots :$

$\to x=\dfrac{15.38139\dots }{3},\:x=-\dfrac{0.77160\dots }{3}$