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

ban oi

ĐK: \(x\ge-3\)

Đặt \(t=\sqrt{x+3}\)  \(\left(t\ge0\right)\)  \(\Rightarrow t^2=x+3\)

\(x^2+2x+\sqrt{x+3}+2x\sqrt{x+3}=9\)

\(x^2+x+\left(x+3\right)+t+2xt=12\)

\(t^2+t\left(2x+1\right)+\left(x^2+x-12\right)=0\)

Goi phương trình trên là phương trình bậc 2 ẩn t 

\(\Delta=\left(2x+1\right)^2-4\cdot1\cdot\left(x^2+x-12\right)\)

\(=4x^2+4x+1-4x^2-4x+48=49>0\)

\(\Rightarrow\)Phương trình có hai nghiệm phân biệt 

\(t_1=\frac{-2x-1-\sqrt{49}}{2\cdot1}=\frac{-2x-8}{2}=-x-4\)

\(t_2=\frac{-2x-1+\sqrt{49}}{2}=3-x\)

+) \(t=-x-4\)

\(\Rightarrow\sqrt{x+3}=-x-4\)

ĐK : \(x\le-4\)

Bình phương 2 vế \(\Rightarrow x+3=x^2+8x+16\)

\(x^2+7x+13=0\)

\(\Delta=-3< 0\Rightarrow x\in\varnothing\)

+) \(t=3-x\)

\(\Rightarrow\sqrt{x+3}=3-x\)

ĐK : \(x\le3\)

BÌnh phương 2 vế \(\Rightarrow x+3=9-6x+x^2\)

\(x^2+7x-6=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-7+\sqrt{73}}{2}\left(tm\right)\\x=\frac{-7-\sqrt{73}}{2}\left(ktm\right)\end{cases}}\)

Vậy \(S=\left\{\frac{-7+\sqrt{73}}{2}\right\}\)

Tớ đã trả lời ở câu hỏi mới nhất r nên xin phép được xóa câu hỏi này nhé

Sửa lại đề bài cho mk là: \(\sqrt{2x+3+\sqrt{x+2}}+\sqrt{2x+2-\sqrt{x+2}}=1+2\sqrt{x+2}\)

\(1.\sqrt{16-8x+x^2}=4-x\)

\(\sqrt{\left(4-x\right)^2}=4-x\)

\(4-x-4+x=0\)

= 0 phương trình vô nghiệm.

\(2.\sqrt{4x^2-12x+9}=2x-3\)

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

\(2x-3-2x+3=0\)

= 0 phương trình vô nghiệm.

a: Ta có: \(\sqrt{16-8x+x^2}=4-x\)

\(\Leftrightarrow\left|4-x\right|=4-x\)

hay \(x\le4\)

b: Ta có: \(\sqrt{4x^2-12x+9}=2x-3\)

\(\Leftrightarrow\left|2x-3\right|=2x-3\)

hay \(x\ge\dfrac{3}{2}\)

a/ \(\sqrt{16-8x+x^2}=4-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\\sqrt{\left(4-x\right)^2}=4-x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\\left|4-x\right|=4-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le4\\\left[{}\begin{matrix}4-x=4-x\left(loại\right)\\4-x=x-4\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x=4\)

Vậy...

b/ \(\sqrt{4x^2-12x+9}=2x-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\\sqrt{\left(2x-3\right)^2}=2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\\left[{}\begin{matrix}2x-3=2x-3\left(loại\right)\\2x-3=3-2x\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy...

\(ĐK:x\in R\)

\(\sqrt{x^2+x+4}+\sqrt{x^2+x+1}=\sqrt{2x^2+2x+9}\) (*)

Đặt \(x^2+x+1=a;a\ge0\)

\(\rightarrow\left\{{}\begin{matrix}x^2+x+4=a+3\\2x^2+2x+9=2a+7\end{matrix}\right.\)

(*) \(\Rightarrow\sqrt{a+3}+\sqrt{a}=\sqrt{2a+7}\)

\(\Leftrightarrow\left(\sqrt{a+3}+\sqrt{a}\right)^2=\left(\sqrt{2a+7}\right)^2\)

\(\Leftrightarrow a+3+a+2\sqrt{a\left(a+3\right)}=2a+7\)

\(\Leftrightarrow2\sqrt{a\left(a+3\right)}=4\)

\(\Leftrightarrow\sqrt{a\left(a+3\right)}=2\)

\(\Leftrightarrow a\left(a+3\right)=4\)

\(\Leftrightarrow a^2+3a-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\left(tm\right)\\a=-4\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow x^2+x+1=1\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) \((tm)\)

Vậy \(S=\left\{0;-1\right\}\)