Lời giải:

a) ĐK: $x\geq 0$

BPT $\Leftrightarrow \sqrt{x+2}(\sqrt{2}-1)\leq \sqrt{x}$

$\Leftrightarrow (3-2\sqrt{2})(x+2)\leq x$

$\Leftrightarrow x(2-2\sqrt{2})\leq 2(2\sqrt{2}-3)$

$\Leftrightarrow x\geq \frac{2(2\sqrt{2}-3)}{2-2\sqrt{2}}=-1+\sqrt{2}$

Vậy BPT có nghiệm $x\geq -1+\sqrt{2}$

b) ĐK: $x\geq 2$ hoặc $x\leq -2$

BPT $\Leftrightarrow (x-5)\sqrt{x^2-4}-(x-5)(x+5)\leq 0$

$\Leftrightarrow (x-5)[\sqrt{x^2-4}-(x+5)]\leq 0$Ta có 2 TH:

TH1: 

\(\left\{\begin{matrix} x-5\geq 0\\ \sqrt{x^2-4}-(x+5)\leq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 5\\ \sqrt{x^2-4}\leq x+5\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 5\\ x^2-4\leq x^2+10x+25\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 5\\ 29\leq 10x\end{matrix}\right.\Leftrightarrow x\geq 5\)

TH2: 

\(\left\{\begin{matrix} x-5\leq 0\\ \sqrt{x^2-4}-(x+5)\geq 0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x\leq 5\\ x^2-4\geq x^2+10x+25 \end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 5\\ -29\geq 10x\end{matrix}\right.\)

 \(\Leftrightarrow \left\{\begin{matrix} x\leq 5\\ x\leq \frac{-29}{10}\end{matrix}\right.\Leftrightarrow x\leq \frac{-29}{10}\)

Kết hợp đkxđ suy ra $x\geq 5$ hoặc $x\leq \frac{-29}{10}$

2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2x-3\ge0\\2x^2-3x+1\ge0\\x^2+2x-3\le2x^2-3x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge1\\x\le\dfrac{1}{2}\end{matrix}\right.\\x^2-5x+4\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge4\\x\le1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x\le-3\\x\ge4\end{matrix}\right.\)

a, ĐKXĐ : \(\left[{}\begin{matrix}x\le-3\\x\ge0\end{matrix}\right.\)

TH1 : \(x\le-3\) ( LĐ )

TH2 : \(x\ge0\)

BPT \(\Leftrightarrow x^2+2x+x^2+3x+2\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge4x^2\)

\(\Leftrightarrow\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge x^2-\dfrac{5}{2}x\)

\(\Leftrightarrow2\sqrt{\left(x+2\right)\left(x+3\right)}\ge2x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x^2+20x+24\ge4x^2-20x+25\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0\le x< \dfrac{5}{2}\\x\ge\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\ge0\)

Vậy \(S=R/\left(-3;0\right)\)

\(\sqrt{x^{ }2-6x+9}=4-x\)
\(\sqrt{\left(x-3\right)^{ }2}=4-x\)
x-3=4-x
x+x=4+3
2x=7
x=\(\dfrac{7}{2}\)

Lời giải:
a.

PT \(\Leftrightarrow \left\{\begin{matrix} 4-x\geq 0\\ x^2-6x+9=(4-x)^2=x^2-8x+16\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 4\\ 2x=7\end{matrix}\right.\Leftrightarrow x=\frac{7}{2}\)

b.

ĐKXĐ: $x\geq \frac{3}{2}$

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

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

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

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

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

a: Ta có: \(\sqrt{x^2-6x+9}=4-x\)

\(\Leftrightarrow\left|x-3\right|=4-x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=4-x\left(x\ge3\right)\\x-3=x-4\left(x< 3\right)\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow2x=7\)

hay \(x=\dfrac{7}{2}\left(nhận\right)\)

Em 2k8 k biết làm có đúng k

ĐKXĐ : \(\left[{}\begin{matrix}x\le-1\\x\ge3\end{matrix}\right.\) 

Bpt \(\Leftrightarrow\left(x-2\right)\left[x+2-\sqrt{x^2-2x-3}\right]\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge2;x+2\ge\sqrt{x^2-2x-3}\left(1\right)\\x\le2;x+2\le\sqrt{x^2-2x-3}\left(2\right)\end{matrix}\right.\)

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

\(\Leftrightarrow6x+7\ge0\)  (Đ với \(x\ge2\) )

(2) có : \(\sqrt{x^2-2x-3}\ge x+2\)

TH1 : x + 2 < 0 <=> \(x< -2\)  ( Bpt luôn đúng ) 

TH2 : \(x+2\ge0\) ; Bp 2 vế rút gọn được : \(6x+7\le0\Leftrightarrow x\le\dfrac{-7}{6}\)

Khi đó : \(-2\le x\le\dfrac{-7}{6}\)

Vậy ... 

a) \(\sqrt{9x}-5\sqrt{x}=6-4\sqrt{x}\)  (đk: \(x\ge0\))

\(\Leftrightarrow3\sqrt{x}-5\sqrt{x}=6-4\sqrt{x}\)

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

\(\Leftrightarrow2\sqrt{x}=6\)

\(\Leftrightarrow\sqrt{x}=3\)

\(\Leftrightarrow\sqrt{x}=\sqrt{9}\)

\(\Leftrightarrow x=9\)(tmđk)

vậy nghiệm của phtrinh là x = 9

b) \(\sqrt{x^2-6x+9}=6\)     (đk: \(x^2-6x+9\ge0\))

bình phương 2 vế, ta được: \(x^2-6x+9=36\)

\(\Leftrightarrow x^2-6x-27=0\)

\(\Leftrightarrow\left(x-9\right)\left(x+3\right)=0\)

\(\Leftrightarrow x=9\)hoặc \(x=-3\)

a) \(\sqrt{9x}-5\sqrt{x}=6-4\sqrt{x}\)

ĐK : x ≥ 0

⇔ \(\sqrt{3^2x}-5\sqrt{x}+4\sqrt{x}=6\)

⇔ \(\left|3\right|\sqrt{x}-\sqrt{x}=6\)

⇔ \(3\sqrt{x}-\sqrt{x}=6\)

⇔ \(2\sqrt{x}=6\)

⇔ \(\sqrt{x}=3\)

⇔ \(x=9\left(tm\right)\)

b) \(\sqrt{x^2-6x+9}=6\)

⇔ \(\sqrt{\left(x-3\right)^2}=6\)

⇔ \(\left|x-3\right|=6\)

⇔ \(\orbr{\begin{cases}x-3=6\\x-3=-6\end{cases}}\)

⇔ \(\orbr{\begin{cases}x=9\\x=-3\end{cases}}\)

c) \(\sqrt{5+2\sqrt{x}}=3\)

ĐK : x ≥ 0

Bình phương hai vế

⇔ \(5+2\sqrt{x}=9\)

⇔ \(2\sqrt{x}=4\)

⇔ \(\sqrt{x}=2\)

⇔ \(x=4\left(tm\right)\)

d) \(\sqrt{2x}< 4\)

ĐK : x ≥ 0

Bình phương hai vế

⇔ \(2x< 16\)

⇔ \(x< 8\)

Kết hợp với ĐK => Nghiệm của bất phương trình là 0 ≤ x < 8