\(B=\dfrac{6^{30}\cdot3^{11}\cdot7-2^{31}\cdot5\cdot3^{40}}{2^{28}\cdot17\cdot3^{38}-6^{27}\cdot3^{11}}\)

\(=\dfrac{2^{30}\cdot3^{41}\cdot7-2^{31}\cdot5\cdot3^{40}}{2^{28}\cdot17\cdot3^{38}-2^{27}\cdot3^{38}}\)

\(=\dfrac{2^{30}\cdot3^{40}\left(3\cdot7-2\cdot5\right)}{2^{27}\cdot3^{38}\left(2\cdot17-1\right)}\)

\(=\dfrac{2^3\cdot3^2\cdot\left(21-10\right)}{34-1}=\dfrac{2^3\cdot3^2\cdot11}{33}=\dfrac{2^3\cdot3^2}{3}=2^3\cdot3=24\)

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????????

Ta có :

\(\left(5xy^2+9xy-x^2y^2\right):\left(-xy\right)\)

\(=\left(-xy\right).\left(-5y-9+xy\right):\left(-xy\right)\)

\(=-5y-9+xy\)

Thay \(x=1,y=2\) vào ta có :

\(-5y-9+xy=\left(-5\right).2-9+1.2=-17\)

M = 2 2 + 3 8 − 18 = 2 2 + 3 4.2 − 9.2 = 2 2 + 6 2 − 3 2 = 5 2

`6\sqrt(2/3)-\sqrt(24)+2\sqrt(3/8)+2\sqrt(1/6)`

`=6. \sqrt6/3 - \sqrt(2^2 .6) + 2. \sqrt(24)/8 + 2. \sqrt6/6`

`=2\sqrt6-2\sqrt6+ \sqrt6/2 + \sqrt6/3`

`=\sqrt6/2+\sqrt6/3`

`=(3\sqrt6+2\sqrt6)/6`

`=(5\sqrt6)/6`

Ta có: \(6\sqrt{\dfrac{2}{3}}-\sqrt{24}+2\sqrt{\dfrac{3}{8}}+2\sqrt{\dfrac{1}{6}}\)

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

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

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