\(=\sqrt{\dfrac{\left(4a-a\sqrt{5}\right)^2}{3^2}}=\dfrac{\left|a\right|\cdot\left(4-\sqrt{5}\right)}{3}\)

`M=sqrt{(3a-1)^2}+2a-3`

`=|3a-1|+2a-3`

`=3a-1+2a-3(do \ a>=1/3)`

`=5a-4`

`N=sqrt{(4-a)^2}-a+5`

`=|4-a|-a+5`

`=a-4-a+5(do \ a>4)`

`=1`

`I=sqrt{(3-2a)^2}+2-7`

`=|3-2a|-5`

`=3-2a-5(do \ a<3/2)`

`=-2-2a`

`K=(a^2-9)/4*sqrt{4/(a-2)^2}`

`=(a^2-9)/4*|2/(a-2)|`

`=(a^2-9)/(2|a-2|)`

Nếu `3>a>2=>|a-2|=a-2`

`=>K=(a^2-9)/(2(a-2))`

Nếu `a<2=>|a-2|=2-a`

`=>K=(a^2-9)/(2(2-a))`

\(M=\left|3a-1\right|+2a-3\)

Mà \(a-\dfrac{1}{3}\ge0\)

\(\Rightarrow M=3a-1+2a-3=5a-4\)

\(N=\left|4-a\right|-a+5\)

Mà \(4-a< 0\)

\(\Rightarrow N=a-4-a+5=1\)

\(I=\left|3-2a\right|-5\)

Mà \(a-\dfrac{3}{2}< 0\)

\(\Rightarrow I=3-2a-5=-2a-2\)

K, Ta có : \(a-3< 0\)

\(\Rightarrow K=\dfrac{2\left(a^2-9\right)}{4\left|a-2\right|}=\dfrac{\left(a-3\right)\left(a+3\right)}{\left|2a-4\right|}\)
 

a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)

\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)

\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)

=-a-1

b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)

\(=\left|3a-5\right|-2a+4\)

\(=5-3a-2a+4\)

=9-5a

c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)

\(=4a-3-\left|2a-1\right|\)

\(=4a-3-2a+1\)

\(=2a-2\)

d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)

\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)

\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)

\(=-a^2\)

\(A=\left|a-3\right|-3a=3-a-3a=3-4a\)

\(B=4a+3-\left|2a-1\right|=4a+3-2a+1=2a+4\)

\(C=\dfrac{4}{a^2-4}\left|a-2\right|=\dfrac{-4\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=\dfrac{-4}{a+2}\)

\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{\left(a+3\right)^2}{16}}=\dfrac{a^2-9}{12}:\dfrac{\left|a+3\right|}{4}=\dfrac{\left(a-3\right)\left(a+3\right).4}{-12\left(a+3\right)}=\dfrac{3-a}{3}\)

\(A=\sqrt{\left(a-3\right)^2}-3a\)

=3-a-3a

=3-4a

a: \(\sqrt[4]{\left(-\dfrac{4}{5}\right)^4}=\left|-\dfrac{4}{5}\right|=\dfrac{4}{5}\)

b: \(\dfrac{\sqrt{4}}{\sqrt{5}}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)

c: \(\left(\sqrt[3]{9}\right)^2=\left(9^{\dfrac{1}{3}}\right)^2=9^{\dfrac{2}{3}}\)

d: \(\sqrt[5]{\sqrt{a}}=\sqrt[5]{a^{\dfrac{1}{2}}}=a^{\dfrac{1}{2}\cdot\dfrac{1}{5}}=a^{\dfrac{1}{10}}\)

e: \(\sqrt[3]{2^6}=\sqrt[3]{\left(2^2\right)^3}=2^2=4\)

\(x=\dfrac{3\sqrt[3]{8-3\sqrt{5}}}{\sqrt[3]{57}}.\sqrt[3]{8+3\sqrt{5}}=\dfrac{3\sqrt[3]{\left(8-3\sqrt{5}\right)\left(8+3\sqrt[]{5}\right)}}{\sqrt[3]{57}}=\sqrt[3]{\dfrac{19}{57}}=\dfrac{1}{\sqrt[3]{3}}\)

\(y=\dfrac{\left(\sqrt[3]{3}+\sqrt[4]{2}\right)\left(\sqrt[3]{3}-\sqrt[4]{2}\right)}{\sqrt[3]{3}+\sqrt[4]{2}}+\dfrac{\left(\sqrt[4]{2}-\sqrt[3]{81}\right)\left(\sqrt[4]{2}+\sqrt[3]{81}\right)}{\sqrt[4]{2}-\sqrt[3]{81}}\)

\(=\sqrt[3]{3}-\sqrt[4]{2}+\sqrt[4]{2}+\sqrt[3]{81}=\sqrt[3]{3}+3\sqrt[3]{3}=4\sqrt[3]{3}\)

\(T=xy=\dfrac{4\sqrt[3]{3}}{\sqrt[3]{3}}=4\)

g: \(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)

=-(căn 5+2)(căn 5-2)

=-(5-4)=-1

h: \(=\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\dfrac{\sqrt{30}}{3}\right)\left(\dfrac{\sqrt{30}}{5}+\sqrt{2}-\dfrac{4}{5}\sqrt{5}\right)\)

=4/5*căn 10+4/3*căn 6-16/15*căn 15+2/5*căn 15+2-4/5*căn 10+30/15+2/3*căn 15-4/3*căn 6

=4