1: ĐKXĐ: x>=-1/2
\(x\left(5x^3+2\right)-2\left(\sqrt{2x+1}-1\right)=0\)
=>\(x\left(5x^3+2\right)-2\cdot\frac{2x+1-1}{\sqrt{2x+1}+1}=0\)
=>\(x\left(5x^3+2\right)-\frac{4x}{\sqrt{2x+1}+1}=0\)
=>\(x\left(5x^3+2-\frac{4}{\sqrt{2x+1}+1}\right)=0\)
=>x=0(nhận)
2: ĐKXĐ: x>=-3
\(\left(x+1\right)^3=\left(x^4+3x^3\right)\cdot\sqrt{x+3}\)
=>\(\left(x+1\right)^3=x^3\left(x+3\right)\cdot\sqrt{x+3}=\left(x\cdot\sqrt{x+3}\right)^3\)
=>\(x+1=x\cdot\sqrt{x+3}\)
=>\(\sqrt{x^3+3x^2}=x+1\)
=>\(\begin{cases}x^3+3x^2=\left(x+1\right)^2\\ x+1\ge0\end{cases}\Rightarrow\begin{cases}x^3+3x^2-x^2-2x-1=0\\ x\ge-1\end{cases}\)
=>\(\begin{cases}x^3+2x^2-2x-1=0\\ x\ge-1\end{cases}\Rightarrow\begin{cases}\left(x-1\right)\left(x^2+x+1\right)+2x\left(x-1\right)=0\\ x\ge-1\end{cases}\)
=>\(\begin{cases}\left(x-1\right)\left(x^2+3x+1\right)=0\\ x\ge-1\end{cases}\)
TH1: x-1=0
=>x=1(nhận)
TH2: \(x^2+3x+1=0\)
=>\(x^2+3x+\frac94=\frac54\)
=>\(\left(x+\frac32\right)^2=\frac54\)
=>\(\left[\begin{array}{l}x+\frac32=\frac{\sqrt5}{2}\\ x+\frac32=-\frac{\sqrt5}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt5-3}{2}\left(nhận\right)\\ x=\frac{-\sqrt5-3}{2}\left(loại\right)\end{array}\right.\)
