$(\sqrt[3]{2x+1}-3)\sqrt{x+1}=x^2 -x-2\sqrt[3]{2x+1} \Leftrightarrow (\sqrt[3]{2x+1}-3)\sqrt{x+1}+2(\sqrt[3]{2x+1}-3)=x^2 -x-6 \Leftrightarrow (\sqrt[3]{2x+1}-3)(\sqrt{x+1}+2)=(x-3)(x+2) \Leftrightarrow (\sqrt[3]{2x+1}-3)(\frac{x-3}{\sqrt{x+1}-2})=(x-3)(x+2)\Leftrightarrow \begin{bmatrix} &x=3 \\ & x+2-\frac{\sqrt[3]{2x+1}-3}{\sqrt{x+1}-2}=0 \end{bmatrix}$
Với $x+2-\frac{\sqrt[3]{2x+1}-3}{\sqrt{x+1}-2}=0$ $\Rightarrow (x+2)(\sqrt{x+1}-2)=\sqrt[3]{2x+1}-3\Leftrightarrow (\sqrt{x+1})^3+\sqrt{x+1}=(\sqrt[3]{2x+1})^3+\sqrt[3]{2x+1}$
Gọi $f(t)=t^3+t \Rightarrow f'(t)=3t^2 +1>0 \Rightarrow \sqrt{x+1}=\sqrt[3]{2x+1}$
Phương trình có 4 nghiệm : $x=3;x=0;x=\frac{1\pm \sqrt{5}}{2}$
hình như thế ... hihi
- Nghiapnh1002 yêu thích