\(\)a)

\(\sqrt{0,01}-\sqrt{0,25}=\sqrt{\frac{1}{100}}-\sqrt{\frac{1}{4}}=\frac{1}{10}-\frac{1}{2}=\frac{1}{10}-\frac{5}{10}=\frac{-4}{10}=\frac{-2}{5}\)

b)

\(0,5.\sqrt{100}-\sqrt{\frac{1}{4}}=0,5.10-\frac{1}{2}=5-\frac{1}{2}=\frac{10}{2}-\frac{1}{2}=\frac{9}{2}\)

ai k mình mình k lại

a) \(\sqrt{0,01}=0,1;\sqrt{0,25}=0,5\)= 0,1-0,5 = -0,4 

b)  = 0,5 x 10 - \(\frac{1}{2}\)= 4,5 . ( Đơn giản nhỉ :) )

a) \(\sqrt{0.01}-\sqrt{0.25}=0,1-0,5=-0,4\)

b)\(0,5.\sqrt{100}-\sqrt{\frac{1}{4}}=0,5.10-0,5=4,5\)

a) \(\sqrt{0,01}-\sqrt{0,25}=\sqrt{\left(0,1\right)^2}-\sqrt{\left(0,5\right)^2}=0,1-0,5=-0,4\)b) \(0,5\sqrt{100}-\sqrt{\frac{1}{4}}=0,5\sqrt{10^2}-\sqrt{\left(\frac{1}{2}\right)^2}=0,5.10-\frac{1}{2}=5-\frac{1}{2}=\frac{9}{2}\)

a) \(\sqrt{0,01}+\sqrt{0,25}=0,1\cdot0,5=0,05\)

b) \(0,5\sqrt{100}-\frac{\sqrt{1}}{4}=0,5\cdot10-\frac{1}{4}=5-\frac{1}{4}=\frac{19}{4}\)

     a) √0,01-√0,25=\(-\frac{2}{5}\)

     b) 0,5.√100-√14 

     =     5         -\(\sqrt{14}\)

     =      5-\(\sqrt{14}\)

\(a,\sqrt{0,01}-\sqrt{0,25}=\sqrt{\frac{1}{100}}-\sqrt{\frac{1}{4}}=\frac{1}{10}-\frac{1}{2}=\frac{1}{10}-\frac{5}{10}=-\frac{4}{10}=-\frac{2}{5}\)

\(b,0,5\sqrt{100}-\sqrt{\frac{1}{4}}=0,5\cdot10-\frac{1}{2}=\frac{5}{10}\cdot10-\frac{1}{2}=5-\frac{1}{2}=\frac{9}{2}\)

1.

0,2 . \(\sqrt{100}\) - \(\sqrt{\dfrac{16}{25}}\)

= 0,2 . 10 - \(\dfrac{4}{5}\)

= 2 - \(\dfrac{4}{5}\)

= \(\dfrac{6}{5}\)

1/ \(0,2.\sqrt{100}-\sqrt{\dfrac{16}{25}}\)

\(=0,2.10-0,8\)

\(=2-0,8=1,2\)

2/ \(\dfrac{2^7.9^3}{6^5.8^2}\)

\(=\dfrac{93312}{497664}=\dfrac{3}{16}=0,1875\)

3/ \(\sqrt{0,01}-\sqrt{0,25}\)

\(=0,1-0,5\)

\(=-0,4\)

4/ \(0,5.\sqrt{100}-\sqrt{\dfrac{1}{4}}\)

\(=0,5.10-0,5\)

\(=5-0,5=4,5\)

5/ \(7.\sqrt{0,01}+2.\sqrt{0,25}\)

\(=7.0,1+2.0,5\)

\(=0,7+1=1,7\)

6/ \(0,5.\sqrt{100}-\sqrt{\dfrac{1}{25}}\)

\(=0,5.10-0,2\)

\(=5-0,2=4,8\)

a) \(A=\dfrac{1}{\sqrt{25}}+\dfrac{\sqrt{49}}{\sqrt{36}}-\dfrac{2}{\sqrt{100}}.\)

\(=\dfrac{1}{5}+\dfrac{7}{6}-\dfrac{1}{5}.\)

\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\dfrac{7}{6}.\)

\(=0+\dfrac{7}{6}=\dfrac{7}{6}.\)

Vậy \(A=\dfrac{7}{6}.\)

b) \(B=\sqrt{\dfrac{0,01}{1,21}}+3.\dfrac{2}{\sqrt{10^2}+2^2+40}-\dfrac{3}{4}.\)

\(=\dfrac{1}{11}+3.\dfrac{2}{10+4+40}-\dfrac{3}{4}.\)

\(=\dfrac{1}{11}+3.\dfrac{1}{37}-\dfrac{3}{4}.\)

\(=\dfrac{1}{11}+\dfrac{1}{9}-\dfrac{3}{4}.\)

\(=\dfrac{36}{396}+\dfrac{44}{396}-\dfrac{297}{296}.\)

\(=-\dfrac{217}{396}.\)

Vậy \(B=-\dfrac{217}{396}.\)