a) \(C=3\cdot\sqrt{25}-3\cdot\sqrt{\frac{1}{9}}\)

\(C=3\cdot5-3\cdot\frac{1}{3}\)

\(C=15-1=14\)

b) \(D=-4\sqrt{\frac{4}{25}}+3\sqrt{0,16}-2\sqrt{0,04}\)

\(D=-4\cdot\frac{2}{5}+3\cdot\frac{2}{5}-2\cdot\frac{1}{5}\)

\(D=\frac{1}{5}\cdot\left(-8+6-2\right)\)

\(D=\frac{1}{5}\cdot\left(-4\right)=-\frac{4}{5}\)

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~

a)
Ta có :

\(\sqrt{0,16}\) + \(\sqrt{\frac{4}{25}}\) = \(\sqrt{\left(0,4^2\right)}\) + \(\sqrt{\left(\frac{2}{5}\right)^2}\) = 0,4 + \(\frac{2}{5}\) = \(\frac{2}{5}+\frac{2}{5}\) = \(\frac{4}{5}\)

b)

Ta có :

\(\sqrt{3\frac{3}{16}}\) - \(\sqrt{0,36}\) = \(\sqrt{\left(\frac{7}{4}\right)^2}\) - \(\sqrt{\left(0,6^2\right)}\) = \(\frac{7}{4}-\frac{3}{5}=\frac{23}{20}\)

\(a)=4\sqrt{0,16-3+4,3}\)\(=4\sqrt{1,46}\)

\(b)=\frac{1}{2}\)

câu c đúng đè ko vậy

a) \(\sqrt{0,16}+\sqrt{0,04}-\sqrt{0,25}\)

= 0,4 + 0,2 - 0,5 

= 0,1

b) \(\sqrt{85^2-84^2}-\sqrt{26^2-24^2}\)

= \(\sqrt{\left(85-84\right)\left(85+84\right)}\) - \(\sqrt{\left(26-24\right)\left(26+24\right)}\)

= \(\sqrt{169}\) - \(\sqrt{2.50}\)

= 13 - 10

= 3 

 Chúc bạn học tốt

a) Ta có: \(\sqrt{0.16}+\sqrt{0.04}-\sqrt{0.25}\)

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

=0,1

b) Ta có: \(\sqrt{85^2-84^2}-\sqrt{26^2-24^2}\)

=13-10

=3

a,

\(\sqrt{0,0004}=0.02\)

\(\sqrt{\frac{16}{81}}=\frac{\sqrt{16}}{\sqrt{81}}=\frac{4}{9}\)

\(\sqrt{25}=5\)

\(\sqrt{0,16}=0,4\)

b,\(\sqrt{\frac{9}{16}}+\sqrt{\frac{25}{9}}\)

= \(\frac{\sqrt{9}}{\sqrt{16}}+\frac{\sqrt{25}}{\sqrt{9}}\)

= \(\frac{3}{4}+\frac{5}{3}\)

=\(\frac{29}{12}\)

a, \(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)\)

\(=15\dfrac{1}{4}.\left(-\dfrac{7}{5}\right)-25\dfrac{1}{4}.\left(-\dfrac{7}{5}\right)\)

\(=-\dfrac{7}{5}.\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right)\)

\(=-\dfrac{7}{5}.\left(-10\right)\)

\(=\dfrac{7}{5}.10\)

\(=\dfrac{7}{1}.2\)

\(=14\)

b, \(\sqrt{0.16}-\sqrt{0.25}\)

\(=0.4-0.5\)

\(=-0.1\)

\(\begin{array}{l}a)\sqrt {1600}  = 40;\\b)\sqrt {0,16}  = 0,4;\\c)\sqrt {2\frac{1}{4}}  = \sqrt {\frac{9}{4}}  = \frac{3}{2}\end{array}\)

\(\text{a)}\sqrt{1600}=40\)

\(\text{b)}\sqrt{0,16}=0,4\)

\(\text{c)}\sqrt{2\dfrac{1}{4}}=\sqrt{\dfrac{9}{4}}=\dfrac{3}{2}\)

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

b/ \(\sqrt{\left(\frac{2}{5}\right)^2\left(x-2\right)^2}=\frac{2}{5}\left|x-2\right|=\frac{2}{5}\left(x-2\right)=\frac{2x}{5}-\frac{4}{5}\)

c/ \(\sqrt{5^2\left(3-a\right)^2}+3=5\left|3-a\right|+3=\left[{}\begin{matrix}18-5a\left(a\le3\right)\\5a-12\left(a\ge3\right)\end{matrix}\right.\)

d/ \(=\frac{1}{2\left(x-5\right)}.6\left|x-5\right|=\frac{3\left|x-5\right|}{x-5}=\left[{}\begin{matrix}3\left(x>5\right)\\-3\left(x< 5\right)\end{matrix}\right.\)