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)

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)

đk: \(-x^4+3x-1\ge0\)

Có \(-\left(x^4+1\right)\le-2x^2\)

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

Áp dụng bunhia có: \(\sqrt{3x-2x^2}+\sqrt{2x^2-3x+2}\le\sqrt{\left(1+1\right)\left(3x-2x^{^2}+2x^2-3x+2\right)}=2\)

\(\Rightarrow\sqrt{-x^4+3x-1}+\sqrt{2x^2-3x+2}\le2\)  (*)

Có: \(x^4-x^2-2x+4=\left(x^4+1\right)-x^2-2x+3\ge2x^2-x^2-2x+3=\left(x-1\right)^2+2\ge2\) (2*)

Từ (*) (2*) dấu = xảy ra khi x=1 (TM)

Vậy x=1

a/ \(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4\)

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

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

Làm nốt

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

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

Làm nốt

a/ \(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=4\)

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

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

Làm nốt

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

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

\(ĐK:x\in R\)

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

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 \(S=\left\{0;-1\right\}\)

a.

ĐKXĐ: \(x\ge1\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x\ge-1\)

\(x^2-6x+9+x+1-4\sqrt{x+1}+4=0\)

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

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

\(\Leftrightarrow x=3\)

c.

ĐKXĐ: \(-2\le x\le\dfrac{4}{5}\)

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

\(VT\le2x+\dfrac{1}{2}\left(9+4-5x\right)+\dfrac{1}{2}\left(1+x+2\right)=8\)

Dấu "=" xảy ra khi và chỉ khi \(x=-1\)

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, ĐKXĐ: ...

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

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

\(\Leftrightarrow3x^2-2x+6=4x^2-12x+9\)

\(\Leftrightarrow4x^2-10x+3=0\)

.....

b, ĐKXĐ: ...

\(\sqrt{x+1}+\sqrt{x-1}=4\\ \Leftrightarrow x+1+x-1+2\sqrt{\left(x+1\right)\left(x-1\right)}=16\\ \Leftrightarrow2\sqrt{x^2-1}=16-2x\\ \Leftrightarrow\sqrt{x^2-1}=8-x\\ \Leftrightarrow x^2-1=64-16x+x^2\\ \Leftrightarrow65-16x=0\\ \Leftrightarrow x=\dfrac{65}{16}\)