h) \(x-2-\sqrt{4-4x+x^2}\)

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

\(=x-2-\left|2-x\right|\)

\(=x-2-2+x\)

\(=2x-4\)

g) \(x-2-\sqrt{4-4x+x^2}\)

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

\(=x-2-\left|2-x\right|\)

\(=x-2-\left[-\left(2-x\right)\right]\)

\(=x-2+2-x\)

\(=0\)

i) \(3-x+\sqrt{9+6x+x^2}\)

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

\(=3-x+\left|3+x\right|\)

\(=3-x-3-x\)

\(=-2x\)

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

Bài 5: 

\(\widehat{B}=60^0\)

\(AB=8\sqrt{3}\left(cm\right)\)

\(BC=16\sqrt{3}\left(cm\right)\)

Bài 1: 

a) \(\dfrac{a+\sqrt{a}}{\sqrt{a}}=\sqrt{a}+1\)

b) \(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}=\dfrac{\left|x-3\right|}{3-x}=\pm1\)

Bài 2: 

a) \(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}=\pm\dfrac{1}{3x+1}\)

b) \(4-x-\sqrt{x^2-4x+4}=4-x-\left|x-2\right|=\left[{}\begin{matrix}6-2x\left(x\ge2\right)\\2\left(x< 2\right)\end{matrix}\right.\)

a) \(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}=\left|3a^2\right|=3a^2\)

b) \(2\sqrt{a^2}-5a=2\left|a\right|-5a=-2a-5a=-7a\)

c) \(\sqrt{16\left(1+4x+4x^2\right)}=\sqrt{\left[4\left(1+2x\right)\right]^2}=\left|4\left(1+2x\right)\right|=4\left(1+2x\right)\)

a.

\(B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\left(x\ge-1\right)\)

\(B=\sqrt{16}.\sqrt{x+1}-\sqrt{9}.\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}+\sqrt{x+1}\)

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

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

\(B=4.\sqrt{x+1}\)

b.

\(B=16\\\)

\(\Rightarrow4\sqrt{x+1}=16\)

\(\Rightarrow\sqrt{x+1}=\dfrac{16}{4}=4\)

\(\Rightarrow x+1=4^2\)

\(\Rightarrow x+1=16\rightarrow x=16-1=15\) (thỏa mãn)

vậy x=15

1.

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

$=x+3+(3-x)=6$

2.

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

$=|x+2|-|x|=x+2-(-x)=2x+2$
3.

$\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}$

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

$=|\sqrt{x^2-1}+1|+|\sqrt{x^2-1}-1|$

$=\sqrt{x^2-1}+1+|\sqrt{x^2-1}-1|$

4.

$\frac{\sqrt{x^2-2x+1}}{x-1}=\frac{\sqrt{(x-1)^2}}{x-1}$

$=\frac{|x-1|}{x-1}=\frac{x-1}{x-1}=1$

5.

$|x-2|+\frac{\sqrt{x^2-4x+4}}{x-2}=2-x+\frac{\sqrt{(x-2)^2}}{x-2}$
$=2-x+\frac{|x-2|}{x-2}|=2-x+\frac{2-x}{x-2}=2-x+(-1)=1-x$

6.

$2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\frac{\sqrt{(x-5)^2}}{x-5}$

$=2x-1-\frac{|x-5|}{x-5}$

a) Ta có: \(A=3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+30\)

\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+30\)

\(=14\sqrt{2x}+30\)

b) Ta có: \(B=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\cdot\sqrt{\dfrac{9x^3}{64}}\)

\(=4\cdot\dfrac{5\sqrt{x}}{2}-\dfrac{8}{3}\cdot\dfrac{3\sqrt{x}}{2}-\dfrac{4}{3x}\cdot\dfrac{3x\sqrt{x}}{8}\)

\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}\)

\(=\dfrac{11}{2}\sqrt{x}\)

c) Ta có: \(\dfrac{y}{2}+\dfrac{3}{4}\sqrt{9y^2-6y+1}-\dfrac{3}{2}\)

\(=\dfrac{1}{2}y+\dfrac{3}{4}\left(1-3y\right)-\dfrac{3}{2}\)

\(=\dfrac{1}{2}y+\dfrac{3}{4}-\dfrac{9}{4}y-\dfrac{3}{2}\)

\(=-\dfrac{7}{4}y-\dfrac{3}{4}\)

a: Ta có: \(x=\sqrt{28-16\sqrt{3}}+2\sqrt{3}\)

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

=4

Thay x=4 vào B, ta được:

\(B=\dfrac{2-4}{2}=-1\)