Câu 1: 

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

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

Trường hợp 1: x<1

(1) trở thành 1-x+2-x=3

=>3-2x=3

=>x=0(nhận)

Trường hợp 2: 1<=x<2

(1) trở thành x-1+2-x=3

=>1=3(loại)

Trường hợp 3: x>=2

(1) trở thành x-1+x-2=3

=>2x-3=3

=>2x=6

hay x=3(nhận)

ĐKXĐ: \(\begin{cases}x+3\ge0\\ 3-2x\ge0\end{cases}\Rightarrow\begin{cases}x\ge-3\\ 2x\le3\end{cases}\Rightarrow-3\le x\le\frac32\)

\(x+4\sqrt{x+3}+2\sqrt{3-2x}=11\)

=>\(x-1+4\sqrt{x+3}-8+2\sqrt{3-2x}-2=0\)

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

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

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

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

=>x-1=0

=>x=1(nhận)

Đặt \(\sqrt{x^2+4}=a>0;\sqrt{x^2+9}=b>0\Rightarrow b^2-a^2=5\)

PT \(\Leftrightarrow\sqrt{x^2+4}+\sqrt{x^2+9}=5\Leftrightarrow a+b=b^2-a^2\Leftrightarrow\left(b-a-1\right)\left(a+b\right)=0\)

...

_{Is that true?}

\(\sqrt{9.\left(x-1\right)^2}-12=0\)

=> 3.(x - 1) - 12 = 0

=> 3x - 15 = 0

=> 3x = 15

=> x = 5

b) \(\sqrt{4.\left(3-x\right)}=16\) (ĐKXĐ: x ≤ 3)

\(\Rightarrow\sqrt{3-x}=8\)

=> 3 - x = 64

=> x = -61

Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}-2\sqrt{16x+16}=\sqrt{x+1}-8\)

\(\Leftrightarrow3\sqrt{x+1}+2\sqrt{x+1}-8\sqrt{x+1}-\sqrt{x+1}=-8\)

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

\(\Leftrightarrow x+1=4\)

hay x=3

ĐKXĐ: \(\frac{x+1}{x-3}\ge0\)

=>x>=3 hoặc x<-1

Ta có: \(\left(x-3\right)\left(x+1\right)+4\cdot\left(x-3\right)\cdot\sqrt{\frac{x+1}{x-3}}=-3\)

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

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

mà \(\left(\sqrt{\left(x-3\right)\left(x+1\right)}+3\right)\left(\sqrt{\left(x-3\right)\left(x+1\right)}+1\right)\ge3\cdot1=3>0\forall x\) thỏa mãn ĐKXĐ
nên x∈∅

\(\sqrt{x-1}+\sqrt{x-4}=x^2-22\)(ĐKXĐ:x>=căn 22)

\(\Leftrightarrow\sqrt{x-1}-2+\sqrt{x-4}-1=x^2-25\)

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

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

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

Vì \(x\ge\sqrt{22}\)nên \(\frac{1}{\sqrt{x-1}+2}-\frac{1}{\sqrt{x-4}+1}-x-5< 0\)

\(\Rightarrow x-5=0\Leftrightarrow x=5\)