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

\(ĐK:x\in R\)

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

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 \(S=\left\{0;-1\right\}\)

Ta có: \(\sqrt{x^2+2x+3}+\sqrt{x^2+x+2}=2x+2\)

Bình phương 2 vế ta có:

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

\(\Leftrightarrow2\sqrt{\left(x^2+2x+3\right)\left(x^2+x+2\right)}=4x^2+8x+4-2x^2-3x-5\)

\(\Leftrightarrow2\sqrt{\left(x^2+2x+3\right)\left(x^2+x+2\right)}=2x^2+5x-1\)​\(\Leftrightarrow2\sqrt{\left(x^2+2x+3\right)\left(x^2+x+2\right)}=2x^2+5x-1\)​

Bình phương 2 vế, ta được:

\(4\left(x^2+2x+3\right)\left(x^2+x+2\right)=\left(2x^2+5x-1\right)^2\) ( ĐK:\(\left[{}\begin{matrix}x\le\dfrac{-5-\sqrt{33}}{4}\\x\ge\dfrac{-5+\sqrt{33}}{4}\end{matrix}\right.\))

\(\Leftrightarrow4\left(x^4+x^3+2x^2+2x^3+2x^2+4x+3x^2+3x+6\right)=4x^4+20x^3+21x^2-10x+1\)

\(\Leftrightarrow4x^4+4x^3+8x^2+8x^3+8x^2+16x+12x^2+12x+24=4x^4+20x^3+21x^2-10x+1\)\(\Leftrightarrow-8x^3+7x^2+38x+23=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{8}\\x=-1\left(loai\right)\end{matrix}\right.\)

Vậy nghiệm của PT là \(x=\dfrac{23}{8}\)

Đặt \(\sqrt{x^2+2x+3}=a;\sqrt{x^2+x+2}=b\) ĐK : \(a;b>0\)

PT <=> a + b = 2(a² - b²) 

<=> a + b = 2(a - b)(a + b)

<=> (a + b)(2a - 2b - 1) = 0

<=> \(\left[{}\begin{matrix}a+b=0\\2a=2b+1\end{matrix}\right.\Leftrightarrow2a=2b+1\left(\text{vì a ; b > 0}\right)\)

Khi đó \(2\sqrt{x^2+2x+3}=2\sqrt{x^2+x+2}+1\)

\(\Leftrightarrow4\left(x^2+2x+3\right)=4\left(x^2+x+2\right)+4\sqrt{x^2+x+2}+1\)

<=> \(4\sqrt{x^2+x+2}=4x+3\)

\(\Leftrightarrow\left\{{}\begin{matrix}16\left(x^2+x+2\right)=16x^2+24x+9\\x\ge-\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x=23\\x\ge-\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow x=\dfrac{23}{8}\)

a.

ĐKXĐ: \(1\le x\le7\)

\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

Biến đổi pt đầu:

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow a^2b^2-b^4=b-a\)

\(\Leftrightarrow b^2\left(a+b\right)\left(a-b\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(b^2\left(a+b\right)+1\right)=0\)

\(\Leftrightarrow a=b\)

\(\Leftrightarrow\sqrt{x}=\sqrt{y-1}\Rightarrow y=x+1\)

Thế vào pt dưới:

\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)

\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)

\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)

\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)

\(\Leftrightarrow...\)

Từ pt suy ra \(x\ge0\).

PT \(\Leftrightarrow\sqrt{2x-2\sqrt{2x-1}}+\sqrt{2x+2\sqrt{2x-1}}=2x\)

\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|+\left|\sqrt{2x-1}+1\right|=2x\). (*)

+) \(\sqrt{2x-1}-1\ge0\Leftrightarrow x\ge1\): Khi đó (*) tương đương \(2\sqrt{2x-1}=2x\Leftrightarrow x^2-2x+1=0\Leftrightarrow x=1\) (thoả mãn)

+) \(\sqrt{2x-1}-1< 0\Leftrightarrow x< 1\): Khi đó (*) tương đương \(2=2x\Leftrightarrow x=1\), vô lí.

Vậy x = 1

Lời giải:

1. ĐKXĐ: $x\geq \frac{-5+\sqrt{21}}{2}$

PT $\Leftrightarrow x^2+5x+1=x+1$

$\Leftrightarrow x^2+4x=0$

$\Leftrightarrow x(x+4)=0$

$\Rightarrow x=0$ hoặc $x=-4$

Kết hợp đkxđ suy ra $x=0$

2. ĐKXĐ: $x\leq 2$

PT $\Leftrightarrow x^2+2x+4=2-x$

$\Leftrightarrow x^2+3x+2=0$

$\Leftrightarrow (x+1)(x+2)=0$

$\Leftrightarrow x+1=0$ hoặc $x+2=0$

$\Leftrightarrow x=-1$ hoặc $x=-2$
3.

ĐKXĐ: $-2\leq x\leq 2$

PT $\Leftrightarrow \sqrt{2x+4}=\sqrt{2-x}$

$\Leftrightarrow 2x+4=2-x$

$\Leftrightarrow 3x=-2$

$\Leftrightarrow x=\frac{-2}{3}$ (tm)

ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\sqrt{2x-5}}=14\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)

\(\Leftrightarrow\left|\sqrt{2x-5}+1\right|+\left|\sqrt{2x-3}+3\right|=14\)

\(\Leftrightarrow2\sqrt{2x-5}=10\)

\(\Leftrightarrow\sqrt{2x-5}=5\)

\(\Leftrightarrow2x-5=25\)

\(\Leftrightarrow x=15\)