a) \(\dfrac{12}{1+\sqrt{5}}+\dfrac{15}{\sqrt{5}}-\dfrac{\sqrt{20}-5}{2-\sqrt{5}}\)

=\(\dfrac{12\left(1-\sqrt{5}\right)}{-4}+\dfrac{15\sqrt{5}}{5}-\dfrac{\left(\sqrt{20}-5\right)\left(2+\sqrt{5}\right)}{-1}\)

=\(-3+3\sqrt{5}-\sqrt{5}+3\sqrt{5}+4\sqrt{5}+10-10-5\sqrt{5}\)

=\(5\sqrt{5}-3\)

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

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

=\(\dfrac{-x+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

1/ Đặt \(\sqrt{x^2+2}=t>0\Rightarrow x^2=t^2-2\)

\(t^2-2+\left(3-t\right)x-1-2t=0\)

\(\Leftrightarrow t^2-2t-3-\left(t-3\right)x=0\)

\(\Leftrightarrow\left(t-3\right)\left(t+1\right)-\left(t-3\right)x=0\)

\(\Leftrightarrow\left(t-3\right)\left(t+1-x\right)=0\)

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

\(\left(1\right)\Leftrightarrow x^2=7\Rightarrow x=\pm\sqrt{7}\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2+2=\left(x-1\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2+2=x^2-2x+1\end{matrix}\right.\) \(\Rightarrow x=\dfrac{-1}{2}\left(l\right)\)

Vậy nghiệm pt là \(x=\pm\sqrt{7}\)

2/

\(x^2+3-6x\sqrt{x^2+3}+9x^2-\sqrt{x^2+3}+3x-2=0\)

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

Đặt \(\sqrt{x^2+3}-3x=t\)

\(\Rightarrow t^2-t-2=0\) \(\Rightarrow\left[{}\begin{matrix}t=-1\\t=2\end{matrix}\right.\)

TH1: \(\sqrt{x^2+3}-3x=-1\Rightarrow\sqrt{x^2+3}=3x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-1\ge0\\x^2+3=\left(3x-1\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\8x^2-6x-2=0\end{matrix}\right.\) \(\Rightarrow x=1\)

TH2: \(\sqrt{x^2+3}-3x=2\Leftrightarrow\sqrt{x^2+3}=3x+2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-2}{3}\\x^2+3=\left(3x+2\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-2}{3}\\8x^2+12x+1=0\end{matrix}\right.\) \(\Rightarrow x=\dfrac{-3+\sqrt{7}}{4}\)

3/ ĐKXĐ: \(\dfrac{3}{2}\le x\le\dfrac{5}{2}\)

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

\(\Rightarrow VT\le2\)

\(VP=3\left(x^2-4x+4\right)+2=3\left(x-2\right)^2+2\ge2\)

\(\Rightarrow VT=VP\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\2x-3=5-2x\end{matrix}\right.\) \(\Rightarrow x=2\)

Vậy pt có nghiệm duy nhất \(x=2\)

4/

ĐKXĐ: \(x\ge\dfrac{-5}{4}\)

\(x^2-2x+1+4x+5-6\sqrt{4x+5}+9=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\\sqrt{4x+5}-3=0\end{matrix}\right.\) \(\Rightarrow x=1\)

Vậy pt có nghiệm duy nhất \(x=1\)

a. 

ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow 6\sqrt{2x}-4\sqrt{2x}+5\sqrt{2x}=21$
$\Leftrightarrow 7\sqrt{2x}=21$

$\Leftrightarrow \sqrt{2x}=3$

$\Leftrightarrow 2x=9$

$\Leftrightarrow x=\frac{9}{2}$ (tm)

b.

ĐKXĐ: $x\geq -2$

PT $\Leftrightarrow \sqrt{25(x+2)}+3\sqrt{4(x+2)}-2\sqrt{16(x+2)}=15$

$\Leftrightarrow 5\sqrt{x+2}+6\sqrt{x+2}-8\sqrt{x+2}=15$

$\Leftrightarrow 3\sqrt{x+2}=15$

$\Leftrightarrow \sqrt{x+2}=5$

$\Leftrightarrow x+2=25$

$\Leftrightarrow x=23$ (tm)

c.

$\sqrt{(x-2)^2}=12$

$\Leftrightarrow |x-2|=12$

$\Leftrightarrow x-2=12$ hoặc $x-2=-12$

$\Leftrightarrow x=14$ hoặc $x=-10$

e.

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

Nếu $x\geq \frac{1}{2}$ thì $2x-1-x=3$

$\Leftrightarrow x=4$ (tm)

Nếu $x< \frac{1}{2}$ thì $1-2x-x=3$

$\Leftrightarrow x=\frac{-2}{3}$ (tm)

2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)

\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)

Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)

\(\Rightarrow x=3\)

c,\(pt\Leftrightarrow3\left(x-1\right)+\frac{x-1}{4x}+\left(2-\sqrt{3x+1}\right)=0\)

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

\(\Rightarrow x=1\)

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

bạn làm nốt pần này nhá

c) \(\sqrt{\left(x-2\right)^2}=10\)

\(x-2=10\)

\(x=12\)

d) \(\sqrt{9x^2-6x+1}=15\)

\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)

\(\sqrt{\left(3x-1\right)^2}=15\)

\(3x-1=15\)

\(3x=16\)

\(x=\dfrac{16}{3}\)

a) \(đk:x\ge0\)

\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)

\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)

b) \(đk:x\ge-2\)

\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)

\(\Leftrightarrow13\sqrt{x+2}=26\)

\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)

c) \(pt\Leftrightarrow\left|x-2\right|=10\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)

\(\Leftrightarrow\left|3x-1\right|=15\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)

e) \(đk:x\ge\dfrac{8}{3}\)

\(pt\Leftrightarrow3x+4=9x^2-48x+64\)

\(\Leftrightarrow9x^2-51x+60=0\)

\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

1.

6x + 1 ≥0

<=>6x≥-1

<=>x≥-1/6

2.

3x - 5 > 0 

<=> 3x > 5

<=> x > 5/3

3.

x - 7 > 0

<=> x > 7

4. 

-3x ≥0

<=>x≤0

Bài 1: 

a: \(=\left|5-\sqrt{3}\right|-\left|\sqrt{3}-2\right|\)

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

b; \(B=\dfrac{\left(2-\sqrt{3}\right)\cdot\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\cdot\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\left(2-\sqrt{3}\right)\cdot\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)

\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-6\sqrt{3}+10-9+5\sqrt{3}}{\sqrt{2}}\)

\(=\dfrac{20-18}{\sqrt{2}}=\sqrt{2}\)

c: \(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

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

\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}=1\)

d: \(A=\left(\sqrt{5}-1\right)\cdot\sqrt{6+2\sqrt{5}}\)

\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)