1. Đk: \(x\ge-\frac{1}{3}\)

\(pt\Leftrightarrow4x^3+5x^2+3x+1-\left(2x+1\right)=\sqrt{3x+1}-\left(2x+1\right)\)

\(\Leftrightarrow x\left(x+1\right)\left(4x+1\right)=\frac{3x+1-\left(2x+1\right)^2}{\sqrt{3x+1}+2x+1}\)

\(\Leftrightarrow x\left(x+1\right)\left(4x+1\right)+\frac{x\left(4x+1\right)}{\sqrt{3x+1}+2x+1}=0\)

\(\Leftrightarrow x\left(4x+1\right)\left(x+1+\frac{1}{\sqrt{3x+1}+2x+1}\right)=0\)

Dễ thấy \(x+1+\frac{1}{\sqrt{3x+1}+2x+1}>0\forall x\ge-\frac{1}{3}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{4}\end{cases}}\)

a) kết quả là 10 + \(\sqrt{2}\) nên ko là số nguyên

câu thứ 2 chịu

\(ĐK:x\ge\frac{-1}{3}\)

\(4x^3+5x^2+1=\sqrt{3x+1}-3x\Leftrightarrow4x^3+5x^2+3x+1-\sqrt{3x+1}=0\)\(\Leftrightarrow x\left(4x^2+5x+3\right)-\frac{3x}{\sqrt{3x+1}+1}=0\)\(\Leftrightarrow x\left(4x^2+5x+3-\frac{3}{\sqrt{3x+1}+1}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tmđk\right)\\4x^2+5x+3-\frac{3}{\sqrt{3x+1}+1}=0\end{cases}}\)

Xét phương trình \(4x^2+5x+3-\frac{3}{\sqrt{3x+1}+1}=0\)\(\Leftrightarrow\left(4x^2+5x+3\right)\sqrt{3x+1}+4x^2+5x=0\)\(\Leftrightarrow\left[\left(x+1\right)\left(4x+1\right)+2\right]\sqrt{3x+1}+4x^2+5x=0\)\(\Leftrightarrow\left(x+1\right)\left(4x+1\right)\sqrt{3x+1}+2\sqrt{3x+1}+4x^2+5x=0\)\(\Leftrightarrow\left(x+1\right)\left(4x+1\right)\sqrt{3x+1}+4x^2+x+4x+1+2\sqrt{3x+1}\)\(-1=0\)

\(\Leftrightarrow\left(x+1\right)\left(4x+1\right)\sqrt{3x+1}+x\left(4x+1\right)+4x+1\)\(+\frac{12x+3}{2\sqrt{3x+1}+1}=0\)

\(\Leftrightarrow\left(4x+1\right)\left[\left(x+1\right)\sqrt{3x+1}+x+1+\frac{3}{2\sqrt{3x+1}+1}\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{4}\left(tmđk\right)\\\left(x+1\right)\sqrt{3x+1}+x+1+\frac{3}{2\sqrt{3x+1}+1}=0\end{cases}}\)

Với \(x\ge\frac{-1}{3}\)thì \(\left(x+1\right)\sqrt{3x+1}+x+1+\frac{3}{2\sqrt{3x+1}+1}>0\)

Vậy phương trình có tập nghiệm \(S=\left\{0;-\frac{1}{4}\right\}\)

ĐK: \(x\ge\frac{-1}{3}\)

\(4x^3+5x^2+1=\sqrt{3x+1}-3x\)

\(\Leftrightarrow4x^3+5x^2+1-\sqrt{3x+1}+3x=0\)

\(\Leftrightarrow4x^3+5x^2+1+\left(2x+1\right)-\sqrt{3x+1}=0\)

\(\Leftrightarrow4x^3+5x^2+x+\frac{\left(2x+1\right)^2-\left(3x+1\right)}{\left(2x+1\right)+\sqrt{3x+1}}=0\)

\(\Leftrightarrow\left(4x^2+x\right)\left(x+1\right)+\frac{4x^2+x}{\left(2x+1\right)+\sqrt{3x+1}}=0\)

\(\Leftrightarrow\left(4x^2+x\right)\left[\left(x+1\right)+\frac{1}{\left(2x+1\right)+\sqrt{3x+1}}\right]=0\)(*)

Với \(x\ge\frac{-1}{3}\)thì \(\left(x+1\right)+\frac{1}{\left(2x+1\right)+\sqrt{3x+1}}>0\)

(*) \(\Leftrightarrow4x^2+x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{4}\end{cases}\left(tmđk\right)}\)

Vậy phương trình có nghiệm \(x=0;x=\frac{-1}{4}\)