a) 1,2+3.1,3=5,1

b) 0,2+2.0,5=1,2

a) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\\\Rightarrow2\sqrt{31}>10\)

Bài 1: 

a) \(\sqrt{1.44}+3\sqrt{1.69}=1.2+3\cdot1.3=1.2+3.9=5.1\)

b) \(\sqrt{0.04}+2\cdot\sqrt{0.25}=0.2+2\cdot0.5=1.2\)

a)\(\sqrt{1}\)+\(\sqrt{9}\)+\(\sqrt{25}\)+\(\sqrt{49}\)+\(\sqrt{81}\)

=1+3+5+7+9

=25

b)=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{4}\)

=\(\dfrac{6}{12}\)+\(\dfrac{4}{12}\)+\(\dfrac{2}{12}\)+\(\dfrac{3}{12}\)

=\(\dfrac{15}{12}\)

c) =0,2+0.3+0,4

= 0.9

d) =9-8+7

=8

j) =1,2-1,3+1.4

= (-0,1)+1,4

=1,4

g) \(\dfrac{2}{5}\)+\(\dfrac{5}{2}\)+\(\dfrac{9}{10}\)+\(\dfrac{3}{4}\)

= (\(\dfrac{4}{10}\)+\(\dfrac{15}{10}\)+\(\dfrac{9}{10}\))+\(\dfrac{3}{4}\)

= \(\dfrac{14}{5}\)+\(\dfrac{3}{4}\)

=\(\dfrac{56}{20}\)+\(\dfrac{15}{20}\)

= \(\dfrac{71}{20}\)

Nhớ tick cho mk nha~

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

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

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

a)

Ta có :

\(\sqrt{\left(-8^2\right)}\) = \(\sqrt{64}\) = 8 vì 8 > 0 và 8² = 64

b)

Ta có :

\(\sqrt{16}\) = 4 vì 4 > 0  và 4² = 16

c)

Ta có :

\(\sqrt{1,44}\) = 1,2 vì 1,2 > 0 và ( 1,2 )² = 1,44

Lời giải:
a. ĐKXĐ: $a\geq 0; a\neq 1$

b.

\(P=\left[\frac{\sqrt{a}(\sqrt{a}+1)}{\sqrt{a}+1}+1\right].\left[\frac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1}-1\right].\frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}-1}\)

\(=(\sqrt{a}+1)(\sqrt{a}-1).\sqrt{2}=\sqrt{2}(a-1)\)

c.

\(P=\sqrt{2}(\sqrt{2+\sqrt{2}}-1)=\sqrt{4+2\sqrt{2}}-\sqrt{2}\)

a. ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{a}\ge0\\\sqrt{a}-1\ne0\\\sqrt{a}+1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\\sqrt{a}\ne1\\\sqrt{a}\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)

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

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

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

a: ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)

b: Ta có: \(P=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\cdot\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-1\right)\cdot\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)

\(=\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)\cdot\sqrt{2}\)

\(=\sqrt{2}a-\sqrt{2}\)

Ta có: \(P=\dfrac{A}{B}\)

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

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

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

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

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

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

Lời giải:

$A=\frac{\sqrt{2}-1}{(1+\sqrt{2})(\sqrt{2}-1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})}+....+\frac{\sqrt{100}-\sqrt{99}}{(\sqrt{99}+\sqrt{100})(\sqrt{100}-\sqrt{99})}$

$=\frac{\sqrt{2}-1}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+....+\frac{\sqrt{100}-\sqrt{99}}{1}$
$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+....+\sqrt{100}-\sqrt{99}$

$=\sqrt{100}-1=10-1=9$