b: ĐKXĐ: 2<=x<=5

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

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

=>\(\frac{3x-3-9}{\sqrt{3x-3}+3}+\frac{1-5+x}{1+\sqrt{5-x}}=\frac{2x-4-4}{\sqrt{2x-4}+2}\)

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

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

=>x-4=0

=>x=4(nhận)

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

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

\(\Leftrightarrow\left|1-x\right|+\left|x-2\right|=3\)

Có: \(VT=\left|1-x\right|+\left|x-2\right|\)

\(\ge\left|1-x+x-2\right|=3=VP\)

Khi \(x=0;x=3\)

b)\(\sqrt{x^2-10x+25}=3-19x\)

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=3-19x\)

\(\Leftrightarrow\left|x-5\right|=3-19x\)

\(\Leftrightarrow x^2-10x+25=361x^2-114x+9\)

\(\Leftrightarrow-360x^2+104x+16=0\)

\(\Leftrightarrow-5\left(5x-2\right)\left(9x+1\right)=0\)

\(\Rightarrow x=\frac{2}{5};x=-\frac{1}{9}\)

c)\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8\sqrt{2x-3}}=5\)

\(\Leftrightarrow\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+8\sqrt{2x-3}+16}=5\)

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

\(\Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}+4\right|=5\)

\(\Leftrightarrow2\sqrt{2x-3}+5=5\)\(\Leftrightarrow\sqrt{2x-3}=0\Leftrightarrow x=\frac{3}{2}\)

c: ĐKXĐ: x>=1/2

Ta có: \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt2\)

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

=>\(\sqrt{2x-1+2\cdot\sqrt{2x-1}\cdot1+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}=2\)

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

=>\(\sqrt{2x-1}+1+\left|\sqrt{2x-1}-1\right|=2\)

=>\(\left|\sqrt{2x-1}-1\right|=2-\sqrt{2x-1}-1=-\sqrt{2x-1}+1=-\left(\sqrt{2x-1}-1\right)\)

=>\(\sqrt{2x-1}-1\le0\)

=>\(\sqrt{2x-1}\le1\)

=>2x-1<=1

=>2x<=2

=>x<=1

=>1/2<=x<=1

d:

ĐKXĐ: x>=-1/4

\(x+\sqrt{x+\frac12+\sqrt{x+\frac14}}=4\)

=>\(x+\sqrt{x+\frac14+2\cdot\sqrt{x+\frac14}\cdot\frac12+\frac14}=4\)

=>\(x+\sqrt{\left(\sqrt{x+\frac14}+\frac12\right)^2}=4\)

=>\(x+\sqrt{x+\frac14}+\frac12=4\)

=>\(x+\frac12+\sqrt{x+\frac14}=4\)

=>\(x+\frac14+2\cdot\sqrt{x+\frac14}\cdot\frac12+\frac14=4\)

=>\(\left(\sqrt{x+\frac14}+\frac12\right)^2=4\)

=>\(\sqrt{x+\frac14}+\frac12=2\)

=>\(\sqrt{x+\frac14}=2-\frac12=\frac32\)

=>\(x+\frac14=\frac94\)

=>x=2(nhận)

a: ĐKXĐ: x>=-2

\(\sqrt{5x+10}=8-x\)

=>\(\begin{cases}8-x\ge0\\ \left(8-x\right)^2=5x+10\end{cases}\Rightarrow\begin{cases}x\le8\\ x^2-16x+64=5x+10\end{cases}\)

=>\(\begin{cases}-2\le x\le8\\ x^2-21x+54=0\end{cases}\Rightarrow\begin{cases}-2\le x\le8\\ \left(x-3\right)\left(x-18\right)=0\end{cases}\)

=>x=3

b: ĐKXĐ: \(4x^2+x-12\ge0\)

=>\(x^2+\frac14x-3\ge0\)

=>\(x^2+2\cdot x\cdot\frac18+\frac{1}{64}-\frac{193}{64}\ge0\)

=>\(\left(x+\frac18\right)^2\ge\frac{193}{64}\)

=>\(\left[\begin{array}{l}x+\frac18\ge\frac{\sqrt{193}}{8}\\ x+\frac18\le-\frac{\sqrt{193}}{8}\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge\frac{\sqrt{193}-1}{8}\\ x\le\frac{-\sqrt{193}-1}{8}\end{array}\right.\)

\(\sqrt{4x^2+x-12}=3x-5\)

=>\(\begin{cases}3x-5\ge0\\ \left(3x-5\right)^2=4x^2+x-12\end{cases}\Rightarrow\begin{cases}3x\ge5\\ 9x^2-30x+25-4x^2-x+12=0\end{cases}\)

=>\(\begin{cases}x\ge\frac53\\ 5x^2-31x+37=0\end{cases}\)

\(\Delta=\left(-31\right)^2-4\cdot5\cdot37=221\) >0

=>Phương trình có hai nghiệm phân biệt là

\(\left[\begin{array}{l}x=\frac{31-\sqrt{221}}{2\cdot5}=\frac{31-\sqrt{221}}{10}\left(loại\right)\\ x=\frac{31+\sqrt{221}}{10}\left(nhận\right)\end{array}\right.\)

\(a,\sqrt{2x+5}=\sqrt{1-x}\)

\(\Rightarrow2x+5=1-x\)

\(2x+x=1-5\)

\(3x=-4\Leftrightarrow x=\frac{-4}{3}\)

Vậy \(S=\left\{-\frac{4}{3}\right\}\)thuộc tập nghiệm của pt trên

a. ĐKXĐ \(x\ge2\)

\(\sqrt{x+3}-3+\sqrt{x-2}-2=0\)

\(\Leftrightarrow\dfrac{x-6}{\sqrt{x+3}+3}+\dfrac{x-6}{\sqrt{x-2}+2}=0\)

\(\Leftrightarrow\left(x-6\right)\left(\dfrac{1}{\sqrt{x+3}+3}+\dfrac{1}{\sqrt{x-2}+2}\right)=0\)

\(\Leftrightarrow x-6=0\Leftrightarrow x=6\)

b.

\(\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\x^2-x-1=\left(1-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^2-x-1=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x=2\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow\) Pt vô nghiệm 

\(a.\sqrt{x+3}=5-\sqrt{x-2}\)

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

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

\(x+3+x-2=25\)

\(2x+1=25\)

\(x=12\)

\(b.\sqrt{x^2-x-1}=1-x\)

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

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

\(x^2-x-1-1+2x-x^2=0\)

\(x-2=0\)

\(x=2\)

\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)

\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)

\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)

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)