2\(\sqrt{\dfrac{16}{3}}\)  - 3\(\sqrt{\dfrac{1}{27}}\) - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{8}{\sqrt{3}}\) - \(\dfrac{3}{3\sqrt{3}}\)  - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{8}{\sqrt{3}}\) - \(\dfrac{1}{\sqrt{3}}\) - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{16}{2\sqrt{3}}\) - \(\dfrac{2}{2\sqrt{3}}\) - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{11}{2\sqrt{3}}\)

= \(\dfrac{11\sqrt{3}}{6}\)

f, 2\(\sqrt{\dfrac{1}{2}}\)- \(\dfrac{2}{\sqrt{2}}\) + \(\dfrac{5}{2\sqrt{2}}\)

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

= \(\dfrac{5}{2\sqrt{2}}\)

= \(\dfrac{5\sqrt{2}}{4}\)

(1 + \(\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\)).(1- \(\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\)) 

= \(\dfrac{\sqrt{3}-1+3-\sqrt{3}}{\sqrt{3}-1}\).\(\dfrac{\sqrt{3}+1-3+\sqrt{3}}{\sqrt{3}+1}\)

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

= \(\dfrac{-4}{3-1}\)

= \(\dfrac{-4}{2}\)

= -2

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

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

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

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

= \(\dfrac{12\sqrt{6}+5\sqrt{6}}{6}\)

= \(\dfrac{17\sqrt{6}}{6}\)

KHO THE

\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)

\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)

\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)

\(1,\sqrt{4\left(a-4\right)^2}\left(dkxd:a\ge4\right)\)

\(=\sqrt{4}.\sqrt{\left(a-4\right)^2}\)

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

\(=2\left(a-4\right)\)

\(=2a-8\)

\(2,\sqrt{9\left(b-5\right)^2}\left(dkxd:b< 5\right)\)

\(=\sqrt{9}.\sqrt{\left(b-5\right)^2}\)

\(=\sqrt{3^2}.\left|b-5\right|\)

\(=3\left(-b+5\right)\)

\(=-3b+15\)

d: \(\Leftrightarrow x+5=5\)

hay x=0

A=(-a - b + c) - (-a-b-c)

A= -a-b+c - (-a)+b+c

A= -a+(-b)+c + a+b+c

A= (-a + a) + (-b+b) + c+c

A=0+0 +c +c

B= -1 + 3 - 5 + 7-9 + 11 -......- 2017+ 2019

B= (-1)+3+(-5)+7+(-9)+11+......+(-2017)+2019

B= [(-1)+3]+[(-5)+7]+[(-9)+11]+......+[(-2017)+2019]

B=   (-2) + (-2) + (-2) +.......+ (-2)

    Tổng B có số số hạng là:

        [ 2019 - 1]:2+1=1010(số hạng)

     Tổng B số cặp là:

            1010:2=505(cặp)

        =>B= (-2) + (-2) + (-2) +.......+ (-2) (505 số hạng)

            B= (-2) . 505

            B=   -1010

            Vậy B = -1010     

- Cảm ơn bạn nhiều 

\(i=120^o-90^o=30^o\)

\(i=i'\Leftrightarrow i'=30^o\)