a) Khi a = 6,5 thì :

\(\sqrt{a^2}\)

= \(\left|a\right|\)

=\(\left|6,5\right|\)

= 6,5

Khi a = -0,1 thì :

\(\sqrt{a^2}\)

= \(\left|a\right|\)

= \(\left|-0,1\right|\)

= \(-\left(-0,1\right)\)

= 0,1

Tuong tu nhu tren :

b) \(\sqrt{a^4}=\sqrt{\left(a^2\right)^2}\)

c)\(\sqrt{a^6}=\sqrt{\left(a^3\right)^2}\)

\(\begin{array}{l}a)2.\sqrt 6 .( - \sqrt 6 )\\ =  - 2.\sqrt 6 .\sqrt 6 \\ =  - 2.{(\sqrt 6 )^2}\\ =  - 2.6\\ =  - 12\\b)\sqrt {1,44}  - 2.{(\sqrt {0,6} )^2}\\ = 1,2 - 2.0,6\\ = 1,2 - 1,2\\ = 0\\c)0,1.{(\sqrt 7 )^2} + \sqrt {1,69} \\ = 0,1.7 + 1,3 \\= 0,7 + 1,3 \\= 2\\d)( - 0,1).{(\sqrt {120} )^2} - \frac{1}{4}.{(\sqrt {20} )^2} \\= ( - 0,1).120 - \frac{1}{4}.20\\ =  - 12 - 5\\ =  - (12 + 5)\\ =  - 17\end{array}\)

a: \(=-2\sqrt{6}\cdot\sqrt{6}=-2\cdot\sqrt{6\cdot6}=-2\cdot6=-12\)

b: \(=1.2-2\cdot0.6=1.2-1.2=0\)

c: \(=0.1\cdot7+1.3=0.7+1.3=2\)

d: \(=-0.1\cdot120-\dfrac{1}{4}\cdot20=-12-5=-17\)

a) \({4^6}.\sqrt {0,1}  = 1295,2689\)

b) \(\sqrt[8]{{2,{1^{18}} + 1}} - \sqrt {2,{1^{12}} + 1}  =  - 80,4632\)

c) \(\frac{{1,{5^3}}}{{\sqrt[3]{{6,8}}}} = 1,7814\)

a: \(=\dfrac{-4}{5}\cdot\left|-\dfrac{5}{4}\right|=\dfrac{-4}{5}\cdot\dfrac{5}{4}=-1\)

b: \(=\left|\left(-2\right)^3\right|=8\)

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

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

e: \(=\left|\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\right|=\dfrac{\sqrt{2}-1}{2}\)

f: \(=\left|\dfrac{1}{10}-\dfrac{\sqrt{10}}{10}\right|=\dfrac{\sqrt{10}-1}{10}\)

Bài 1: 

a: \(=\sqrt{225}=15\)

b: \(=\sqrt{\dfrac{2}{5}\cdot\dfrac{32}{5}}=\sqrt{\dfrac{64}{25}}=\dfrac{8}{5}\)

c: \(=\sqrt{121\cdot36}=11\cdot6=66\)

d: \(=7\cdot1.2\cdot5=35\cdot1.2=42\)

g: \(=\sqrt{\dfrac{27}{10}\cdot\dfrac{3}{2}\cdot5}=\sqrt{\dfrac{81}{20}\cdot5}=\sqrt{\dfrac{81}{4}}=\dfrac{9}{2}\)

Bài 2: 

a: \(=\dfrac{1}{3}\cdot0.8\cdot8=\dfrac{8}{3}\cdot\dfrac{4}{5}=\dfrac{32}{15}\)

b: \(=\sqrt{\dfrac{100}{9}}=\dfrac{10}{3}\)

c: \(=\sqrt{\dfrac{1}{144}\cdot\dfrac{100}{49}}=\dfrac{1}{12}\cdot\dfrac{10}{7}=\dfrac{5}{6\cdot7}=\dfrac{5}{42}\)

b) \(B=\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right):\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(B=\left[\dfrac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{a}+\sqrt{b}}\right]:\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(B=\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\right]:\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(B=\left(a-\sqrt{ab}+\sqrt{b}\right):\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(B=\dfrac{a-\sqrt{ab}+b}{a-b}+\dfrac{2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(B=\dfrac{a-\sqrt{ab}+b}{a-b}+\dfrac{2\sqrt{ab}-2b}{a-b}\)

\(B=\dfrac{a-\sqrt{ab}+b+2\sqrt{ab}-2b}{a-b}\)

\(B=\dfrac{a+\sqrt{ab}-b}{a-b}\)

a) \(\sqrt{2}A=\sqrt{2x-2\sqrt{x-2}.\sqrt{x+2}}+\sqrt{2x+2\sqrt{x-2}.\sqrt{x+2}}\) (\(x\ge2\) )

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

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

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

\(=\sqrt{x+2}-\sqrt{x-2}+\sqrt{x+2}+\sqrt{x-2}\) ( do \(x+2>x-2\ge0\Leftrightarrow\sqrt{x+2}>\sqrt{x-2}\) )

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

\(\Leftrightarrow A=\sqrt{2}.\sqrt{x+2}\)

Vậy...

b) \(B=\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right):\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\) 

\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}.\dfrac{1}{a-b}+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\dfrac{a-\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\dfrac{a-\sqrt{ab}+b+2\sqrt{ab}-2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\dfrac{a+\sqrt{ab}-b}{a-b}\)

Vậy...

a: \(=\dfrac{-4}{5}\cdot\dfrac{5}{4}=-1\)

b: =8

c: \(=2-\sqrt{3}\)

d: \(=3-2\sqrt{2}\)

e: \(=\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\)

1. với a=2,5 thì \(\sqrt{a^2}\) =\(\left|a\right|=\)\(\left|2.5\right|=2.5\)

với a=0,3 thì \(\sqrt{a^2}\) =\(\left|a\right|=\)\(\left|0,3\right|=0,3\)

với a=-0,1 thì \(\sqrt{a^2}\) =\(\left|a\right|=\)\(\left|-0,1\right|=0,1\)

Bài 4: 

a: \(=\sqrt{\dfrac{10.8}{0.3}}=\sqrt{36}=6\)

b: \(=\sqrt{\dfrac{7}{175}}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\)

c: \(=\sqrt{\dfrac{2.84}{0.71}}=2\)

d: \(=\sqrt{\dfrac{625}{144}}=\dfrac{25}{12}\)