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

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

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

Lời giải:

a. ĐKXĐ: $x\geq 4$

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

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

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

$\Leftrightarrow  \sqrt{x-4}+2=2$

$\Leftrightarrow \sqrt{x-4}=0$

$\Leftrightarrow x=4$ (tm)

b. ĐKXĐ: $x\in\mathbb{R}$

PT $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$

$\Leftrightarrow |2x-1|=|x-3|$

\(\Rightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)

c.

PT \(\Rightarrow \left\{\begin{matrix} 2x-1\geq 0\\ 2x^2-2x+1=(2x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x(x-1)=0\end{matrix}\right.\Rightarrow x=1\)

\(\Leftrightarrow2x^2+2+2\sqrt{\left(x^2+x+1\right)\left(x^2-x+1\right)}=2x^2+4\)

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

\(\Leftrightarrow x^4+x^2=0\)

\(\Leftrightarrow x=0\)

`\sqrt{x^2+x+1}+\sqrt{x^2-x+1}=\sqrt{2x^2+4}`

`<=>2x^2+2+2\sqrt{x^4+x^2+1}=2x^2+3`

`<=>\sqrt{x^4+x^2+1}=1`

`<=>x^4+x^2=0`

`<=>x=0`

a. \(\sqrt[3]{1-2x}+3=0\left(ĐK:x\le\dfrac{1}{2}\right)\)

<=> \(\sqrt[3]{1-2x}=-3\)

<=> \(1-2x=\left(-3\right)^3\)

<=> \(1-2x=-27\)

<=> \(-2x=-28\)

<=> \(x=14\left(TM\right)\)

\(\Leftrightarrow\left(\sqrt[3]{x+1}-1\right)+\left(\sqrt{2x+4}-2\right)< -x\sqrt{2}\)

=>\(\dfrac{x+1-1}{\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1}+\dfrac{2x+4-4}{\sqrt{2x+4}+2}+x\sqrt{2}< 0\)

=>x<0

=>-1<x<0

ĐKXĐ: \(x\ge\dfrac{3}{2}\).

PT đã cho tương đương:

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\Leftrightarrow x=4\left(TMĐK\right)\\\sqrt{2x-3}+\sqrt{x+1}=1\left(1\right)\end{matrix}\right.\).

Ta có \(\left(1\right)\Leftrightarrow2x-3+x+1+2\sqrt{\left(2x-3\right)\left(x+1\right)}=1\)

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

Do đó 3 - 3x \(\ge0\Leftrightarrow x\le1\) (trái với đkxđ).

Suy ra (1) vô nghiệm.

Vậy ncpt là x = 4.

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

ĐKXĐ: \(x\ge\dfrac{1}{2}\)

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

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

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

\(\Leftrightarrow x-1=0\)( vì \(\dfrac{1}{\sqrt{x+3}+2}+\dfrac{2}{\sqrt{2x-1}+1}+1\)>0)

\(\Leftrightarrow x=1\)(thỏa mãn)

Vậy phương trình có nghiệm là x=1