a: \(=\dfrac{3}{4}-\dfrac{5}{6}+\dfrac{3}{2}=\dfrac{9-10+18}{12}=\dfrac{17}{12}\)

b: \(=\left(\dfrac{1}{9}+\dfrac{6}{9}\right)^2-\dfrac{1}{3}=\dfrac{49}{81}-\dfrac{27}{81}=\dfrac{22}{81}\)

c; \(=\dfrac{5}{11}\left(-\dfrac{3}{7}-\dfrac{5}{7}\right)+\dfrac{-8}{7}\cdot\dfrac{6}{11}=\dfrac{-8}{7}\left(\dfrac{5}{11}+\dfrac{6}{11}\right)=-\dfrac{8}{7}\)

d: \(=\dfrac{2^{26}}{2^{15}\cdot2^{12}}=\dfrac{1}{2}\)

a) 9
b)90
c)8
d)7/10
e)2/5

a=9

b=90

c=2

d=0,7

e=0,2

`A=\sqrt{25/[3-2\sqrt{2}]}`

`A=\sqrt{[25(3+2\sqrt{2})]/[9-8]}`

`A=5\sqrt{3+2\sqrt{2}}`

___________

`B=\sqrt{[a^4 b^3]/[a^2 b-ab]}`       `(a > 1;b > 0)`

`B=\sqrt{[ab.a^3 b^2]/[ab(a-1)]}`

`B=ab\sqrt{a/[a-1]}`

A: \(A=\sqrt{9}-3\sqrt{\dfrac{50}{9}}+3\sqrt{8}-\sqrt[3]{27}\)

\(=3-3\cdot\dfrac{5\sqrt{2}}{3}+6\sqrt{2}-3\)

\(=-5\sqrt{2}+6\sqrt{2}=\sqrt{2}\)

b: \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}-1}-6\cdot\sqrt{\dfrac{16}{3}}\)

\(=\left|2-\sqrt{3}\right|+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}-6\cdot\dfrac{4}{\sqrt{3}}\)

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

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

\(A=\sqrt{9}-3\sqrt{\dfrac{50}{9}}+3\sqrt{8}-\sqrt[3]{27}\\ =3-3\cdot\dfrac{1}{3}\sqrt{25\cdot2}+3\sqrt{4\cdot2}-3\\ =3-1\cdot5\sqrt{2}+3\cdot2\sqrt{2}-3\\ =3-5\sqrt{2}+6\sqrt{2}-3\\ =\sqrt{2}\)

\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}-1}-6\sqrt{\dfrac{16}{3}}\\ =\left|2-\sqrt{3}\right|+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}-6\cdot\dfrac{4\sqrt{3}}{3}\\ =2-\sqrt{3}+\sqrt{3}+1-8\sqrt{3}\\ =3-8\sqrt{3}\)

a) Ta có: \(\dfrac{-4}{3}\cdot\sqrt{\left(-0.4\right)^2}\)

\(=-\dfrac{4}{3}\cdot0.4\)

\(=\dfrac{-1.6}{3}=-\dfrac{8}{15}\)

b) Ta có: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}\)

\(=\sqrt[3]{\dfrac{27}{64}}=\dfrac{3}{4}\)

c) Ta có: \(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\)

\(=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{7}\)

\(=\dfrac{6}{7}\)

\(\dfrac{3}{14}\cdot\left(-49\right)=-\dfrac{21}{2}\)

\(\dfrac{3}{4}\cdot\dfrac{-18}{25}=-\dfrac{27}{50}\)

\(1\dfrac{2}{3}\cdot3\dfrac{2}{9}=\dfrac{29}{9}\cdot\dfrac{5}{3}=\dfrac{145}{27}\)

\(40\%\cdot\dfrac{20}{9}=\dfrac{40}{100}\cdot\dfrac{20}{9}=\dfrac{40}{45}=\dfrac{8}{9}\)

a, \(-49.\dfrac{3}{14}=-\dfrac{21}{2}\)

b, \(\dfrac{-18}{25}.\dfrac{3}{4}=-\dfrac{27}{50}\)

c, \(3\dfrac{2}{9}.1\dfrac{2}{3}=\dfrac{29}{9}.\dfrac{5}{3}=\dfrac{145}{27}\)

d, \(\dfrac{20}{9}.40\%=\dfrac{20}{9}.\dfrac{40}{100}=\dfrac{8}{9}\)

a)2

b)-0,4

c)7

a) \(\sqrt{25-9}\) = \(\sqrt{16}\) = 4

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

c)\(\sqrt{2.2^2+4^2}+5^2\) = \(\sqrt{2.4+16+25}\) = \(\sqrt{8+16+25}\) = \(\sqrt{49}\) = 7

a: =(2căn 3-8căn 3)(căn 3-1)

=-6căn 3*(căn 3-1)

=-18+6căn 3

b: \(=\dfrac{6-2\sqrt{5}}{\sqrt{5}-3}-\sqrt{5}+2\)

=-2-căn 5+2=-căn 5

c: \(=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-\dfrac{1}{4}\cdot8\sqrt{2a}\)

=\(3\sqrt{2a}-3a\cdot\sqrt{2a}\)

a,\(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5=\left(\sqrt{\dfrac{25}{16}}-\dfrac{3}{4}\right):5=\left(\dfrac{5}{4}-\dfrac{3}{4}\right):5\)

\(=\dfrac{1}{2}:5=\dfrac{1}{10}\)

b,\(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2=\left[\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\right]^2\)

\(=\left[3-4\right]^2=1\)

c,\(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)=11^2-\left(4\sqrt{3}\right)^2\)

\(=121-48=73\)

d,\(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)

\(=2-2\sqrt{2}+1-3+\dfrac{4\sqrt{2}}{5}+\sqrt{\dfrac{36}{25}.2}\)

\(=-2\sqrt{2}+\dfrac{4\sqrt{2}+6\sqrt{2}}{5}\)

\(=-2\sqrt{2}+\dfrac{10\sqrt{2}}{5}=-2\sqrt{2}+2\sqrt{2}=0\)

e,\(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)

\(=\left(1+\sqrt{2021}\right)\sqrt{2021-2\sqrt{2021}.1+1}\)

\(=\left(1+\sqrt{2021}\right)\sqrt{\left(\sqrt{2021}-1\right)^2}\)

\(=\left(1+\sqrt{2021}\right)\left(\sqrt{2021}-1\right)\)

\(=\sqrt{2021}-1+\sqrt{2021^2}-\sqrt{2021}=2020\)

Bài 1: 

a) \(\sqrt{72}:\sqrt{8}=\sqrt{72:8}=3\)

b) \(\left(\sqrt{28}-\sqrt{7}+\sqrt{112}\right):\sqrt{7}=5\sqrt{7}:\sqrt{7}=5\)

Bài 2: 

a) \(\sqrt{\dfrac{49}{8}}:\sqrt{3\dfrac{1}{8}}=\sqrt{\dfrac{49}{8}:\dfrac{25}{8}}=\sqrt{\dfrac{49}{25}}=\dfrac{7}{5}\)

b) \(\sqrt{54x}:\sqrt{6x}=\sqrt{54x:6x}=\sqrt{9}=3\)

c) \(\sqrt{\dfrac{1}{125}}\cdot\sqrt{\dfrac{32}{35}}:\sqrt{\dfrac{56}{225}}\)

\(=\dfrac{\sqrt{5}}{25}\cdot\dfrac{4\sqrt{2}}{\sqrt{35}}:\dfrac{2\sqrt{14}}{15}\)

\(=\dfrac{\sqrt{5}\cdot4\sqrt{2}\cdot15}{25\cdot\sqrt{35}\cdot\sqrt{14}\cdot2}\)

\(=\dfrac{6}{35}\)