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

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

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

a)\(5,75.\frac{{ - 8}}{9} =\frac{{575}}{100}.\frac{{ - 8}}{9}= \frac{{23}}{4}.\frac{{ - 8}}{9} = \frac{{ - 46}}{9}\)

b)\(2\frac{3}{8}.\left( { - 0,4} \right) = \frac{{19}}{8}.\frac{{ - 4}}{10} =\frac{{19}}{8}.\frac{{ - 2}}{5} = \frac{{ - 19}}{{20}}\);            

c)\(\frac{{ - 12}}{5}:\left( { - 6,5} \right) = \frac{{ - 12}}{5}:\frac{{ - 65}}{10} =\frac{{ - 12}}{5}:\frac{{ - 13}}{2} = \frac{{ - 12}}{5}.\frac{{ - 2}}{{13}} = \frac{{24}}{{65}}\).

b: \(=16-2\cdot4\cdot2\sqrt{5}+20-9-4\sqrt{5}\)

=27-20căn 5

a: 2-4căn 3<0

nên biểu thức ko có giá trị

\(b,\left(4-2\sqrt{5}\right)^2-\left(\sqrt{5}+2\right)^2\\ =\left[\left(4-2\sqrt{5}\right)-\left(\sqrt{5}+2\right)\right].\left[\left(4-2\sqrt{5}\right)+\left(\sqrt{5}+2\right)\right]=\left(2-3\sqrt{5}\right)\left(6-\sqrt{5}\right)\)

`b)(4-2\sqrt5)^2-(\sqrt5+2)^2`

`=(4-2\sqrt5-\sqrt5-2)(4-2\sqrt5+\sqrt5+2)`

`=(2-3\sqrt5)(6-\sqrt5)`

$---------$

Áp dụng HĐT :

`a^2-b^2=(a-b)(a+b)`

\(i,\sqrt{12,1.360}=\sqrt{12,1}.6\sqrt{10}=6.\sqrt{12,1.10}=6.\sqrt{121}=6.\sqrt{11^2}=6.11=66\)

\(k,\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4.6,4}=\sqrt{\dfrac{64}{25}}=\dfrac{\sqrt{8^2}}{\sqrt{5^2}}=\dfrac{8}{5}\)

\(l,-0,4.\sqrt{\left(-0,4\right)^2}=-0,4.0,4=-0,16\)

\(m,\sqrt{2^4.\left(-7\right)^2}=\sqrt{4^2}.\sqrt{\left(-7\right)^2}=4.7=28\)

i, \(\sqrt{12,1\cdot360}=\sqrt{4356}=\sqrt{66^2}=66\)

k, \(\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4\cdot6,4}=\sqrt{\dfrac{64}{25}}=\sqrt{\dfrac{2^6}{5^2}}=\dfrac{2^3}{5}=\dfrac{8}{5}\)

l, \(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\cdot\left|-0,4\right|=-0,4\cdot0,4=-\dfrac{4}{25}\)

m, \(\sqrt{2^4\cdot\left(-7\right)^2}=2^2\cdot\left|-7\right|=4\cdot7=28\)

i)        \(\sqrt{12,1.360}=\sqrt{12,1}.\sqrt{360}=\sqrt{12,1}.\sqrt{36}.\sqrt{10}=\left(\sqrt{12,1}.\sqrt{10}\right)\sqrt{36}=\sqrt{121}.\sqrt{36}=11.6=66\)

k)       \(\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4.6,4}=\sqrt{2,56}=1,6\)

l)        \(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\left|-0,4\right|=-0,4.0,4=0,16\)

m)         \(\sqrt{2^4.\left(-7\right)^2}=\sqrt{2^2.2^2.\left(-7\right)^2}=\left|2.2.7\right|=28\)

\(5\sqrt{\left(-2\right)^4}=5\sqrt{4^2}=5.4=20\)

\(-4\sqrt{\left(-3\right)^6}=-4\sqrt{27^2}=-4.27=-108\)

\(\sqrt{\sqrt{\left(-5\right)^8}}=\sqrt{\sqrt{\left(5^4\right)^2}}=\sqrt{5^4}=\sqrt{25^2}=25\)

a)

\(\begin{array}{l}0,75 - \frac{5}{6} + 1\frac{1}{2} = \frac{3}{4} - \frac{5}{6} + \frac{3}{2}\\ = \frac{9}{{12}} - \frac{{10}}{{12}} + \frac{{18}}{{12}} = \frac{{17}}{{12}}\end{array}\)                                    

b)

\(\begin{array}{l}\frac{3}{7} + \frac{4}{{15}} + \left( {\frac{{ - 8}}{{21}}} \right) + \left( { - 0,4} \right) = \frac{3}{7} + \frac{4}{{15}} - \frac{8}{{21}} - \frac{2}{5}\\ = \left( {\frac{3}{7} - \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} - \frac{2}{5}} \right)\\ = \left( {\frac{9}{{21}} - \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} - \frac{6}{{15}}} \right)\\ = \frac{1}{{21}} + \left( {\frac{{ - 2}}{{15}}} \right)\\ = \frac{5}{{105}} - \frac{{14}}{{105}}\\ = \frac{{ - 9}}{{105}} = \frac{{ - 3}}{{35}}\end{array}\)

c)

\(\begin{array}{l}0,625 + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} + \left( {\frac{{ - 5}}{7}} \right) + 1\frac{2}{3}\\ = \frac{5}{8} + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} - \frac{5}{7} + \frac{5}{3}\\ = \left( {\frac{5}{8} + \frac{3}{8}} \right) + \left( {\frac{{ - 2}}{7} - \frac{5}{7}} \right) + \frac{5}{3}\\ = 1 - 1 + \frac{5}{3} = \frac{5}{3}\end{array}\)          

 d)

\(\begin{array}{l}\left( { - 3} \right).\left( {\frac{{ - 38}}{{21}}} \right).\left( {\frac{{ - 7}}{6}} \right).\left( { - \frac{3}{{19}}} \right)\\ = \frac{{ - 3.\left( { - 38} \right).\left( { - 7} \right).\left( { - 3} \right)}}{{21.6.19}}\\ = \frac{{3.38.7.3}}{{21.6.19}}\\ = \frac{{3.2.19.7.3}}{{3.7.3.2.19}}\\ = 1\end{array}\)

e)

 \(\begin{array}{l}\left( {\frac{{11}}{{18}}:\frac{{22}}{9}} \right).\frac{8}{5} = \left( {\frac{{11}}{{18}}.\frac{9}{{22}}} \right).\frac{8}{5}\\ = \frac{{11.9.4.2}}{{9.2.2.11.5}} = \frac{2}{5}\end{array}\)                                   

 g)

\(\left[ {\left( {\frac{{ - 4}}{5}} \right).\frac{5}{8}} \right]:\left( {\frac{{ - 25}}{{12}}} \right) = \frac{{ - 20}}{{40}}:\left( {\frac{{ - 25}}{{12}}} \right)\\ = \frac{{ - 1}}{2}.\frac{{ - 12}}{{25}} = \frac{6}{{25}}\)

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

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

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

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

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

\(a,=\sqrt{5}\left(2\sqrt{5}-3\right)+3\sqrt{5}=10-3\sqrt{5}+3\sqrt{5}=10\\ b,=5-\sqrt{3}-\left(2-\sqrt{3}\right)=3\\ c,=\dfrac{2\left(\sqrt{5}-1\right)}{4}-\dfrac{2\left(3+\sqrt{5}\right)}{4}=\dfrac{2\sqrt{5}-2-6-2\sqrt{5}}{4}=\dfrac{-8}{4}=-2\)

a) Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

\(=\dfrac{-2\left(\sqrt{3}-\sqrt{8}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{6}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)

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

b) Ta có: \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)

\(=1-\left(\sqrt{2}+\sqrt{3}\right)^2\)

\(=1-5-2\sqrt{6}\)

\(=-4-2\sqrt{6}\)

bài 1: 

a: Ta có: \(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)

\(=6\sqrt{2}-45\sqrt{2}+6\sqrt{2}\)

\(=-33\sqrt{2}\)

b: Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)

\(=10-2\sqrt{21}+14\sqrt{21}\)

\(=12\sqrt{21}+10\)

Bài 2: 

a: Ta có: \(\sqrt{\left(2x+3\right)^2}=8\)

\(\Leftrightarrow\left|2x+3\right|=8\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)

b: Ta có: \(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)

\(\Leftrightarrow4\sqrt{x}=8\)

hay x=4

c: Ta có: \(\sqrt{9x-9}+1=13\)

\(\Leftrightarrow3\sqrt{x-1}=12\)

\(\Leftrightarrow x-1=16\)

hay x=17