a, \(x=0\) không là nghiệm của phương trình
Xét \(x\ne0\), phương trình tương đương:
\(x+\sqrt[3]{x-\dfrac{1}{x}}=2+\dfrac{1}{x}\)
\(\Leftrightarrow x-\dfrac{1}{x}+\sqrt[3]{x-\dfrac{1}{x}}-2=0\)
\(\Leftrightarrow t^3+t-2=0\left(t=\sqrt[3]{x-\dfrac{1}{x}}\right)\)
\(\Leftrightarrow\left(t-1\right)\left(t^2+t+2\right)=0\)
\(\Leftrightarrow t=1\) (Vì \(t^2+t+2>0\))
\(\Leftrightarrow\sqrt[3]{x-\dfrac{1}{x}}=1\)
\(\Leftrightarrow x-\dfrac{1}{x}=1\)
\(\Leftrightarrow x^2-x-1=0\)
\(\Leftrightarrow x=\dfrac{1\pm\sqrt{5}}{2}\)
Vậy ...
b, ĐK: \(x\ge-1\)
Đặt \(\sqrt{x+1}=a\left(a\ge0\right);\sqrt{x^2-x+1}=b\left(b\ge\dfrac{\sqrt{3}}{2}\right)\)
\(2\left(x^2+2\right)=5\sqrt{x^3+1}\)
\(\Leftrightarrow2\left(x+1+x^2-x+1\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\Leftrightarrow2\left(a^2+b^2\right)=5ab\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\)
TH1: \(a=2b\Leftrightarrow\sqrt{x+1}=2\sqrt{x^2-x+1}\)
\(\Leftrightarrow4x^2-5x+3=0\)
\(\Rightarrow\) Vô nghiệm
TH2: \(2a=b\Leftrightarrow2\sqrt{x+1}=\sqrt{x^2-x+1}\)
\(\Leftrightarrow x^2-5x-3=0\)
\(\Leftrightarrow x=\dfrac{5\pm\sqrt{37}}{2}\)
Vậy ...
