\(\sqrt{x^2+2x+1}+\sqrt{x^2-6x+9}\)

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

\(=\left|x+1\right|+\left|x-3\right|\)

\(=x+1+x-3\)

\(=2x-2\)

\(=2\left(x-1\right)\)

b sai \(\sqrt{x^2+6x+9}\) ms đúng

a)\(\sqrt{9x^2}=2x+1\)

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

<=>3|x|=2x+1

dễ r`

Câu a dễ r câu b nha

a, ĐK: ​\(\left(x+1\right)\left(x^2+2x-1\right)\ge0\)​

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

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

TH1: \(x\ge-1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-1}=\sqrt{x+1}\\\sqrt{x^2+2x-1}=3\sqrt{x+1}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-1=x+1\\x^2+2x-1=9x+9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2-7x-10=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

TH2: \(x< -1\)

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

\(\Leftrightarrow...\)

Bài này dài nên ... cho nhanh nha, đoạn sau dễ rồi

x= 0.761322463768116,

x= 0.369494467346496,

x=1.57660410301179

Đúng đề chưa vậy