Question

Giải phương trình:

\(\sqrt[3]{2x+3}\) + 1= \(x^3+3x^2+2x\)

Answer 1

Lời giải:
ĐK:...........

\(\Leftrightarrow x^3+3x^2+2x-\sqrt[3]{2x+3}-1=0\)

\(\Leftrightarrow x^3+3x^2+x-2+(x+1)-\sqrt[3]{2x+3}=0\)

\(\Leftrightarrow x^3+3x^2+x-2+\frac{(x+1)^3-(2x+3)}{(x+1)^2+(x+1)\sqrt[3]{2x+3}+\sqrt[3]{(2x+3)^2}}=0\)

\(\Leftrightarrow (x^3+3x^2+x-2)\left(1+\frac{1}{(x+1)^2+(x+1)\sqrt[3]{2x+3}+\sqrt[3]{(2x+3)^2}}\right)=0\)

Dễ thấy \(\left(1+\frac{1}{(x+1)^2+(x+1)\sqrt[3]{2x+3}+\sqrt[3]{(2x+3)^2}}\right)>0\)

Do đó \(x^3+3x^2+x-2=0\)

\(\Leftrightarrow x^2(x+2)+(x^2+x-2)=0\)

\(\Leftrightarrow x^2(x+2)+(x+2)(x-1)=0\)

\(\Leftrightarrow (x+2)(x^2+x-1)=0\Rightarrow \left[\begin{matrix} x=-2\\ x=\frac{-1\pm \sqrt{5}}{2}\end{matrix}\right.\) (đều thỏa mãn)

Vậy........