MaxWong

You don't need a common denominator though......

\(7\dfrac{5}{6}\times 11\dfrac{2}{4}\\ =7\dfrac{5}{6}\times 11\dfrac{1}{2}\\ =\dfrac{47}{6}\times \dfrac{23}{2}\\ =\dfrac{47\times 23}{6\times 2}\\ =\dfrac{1081}{12}\\ =90\dfrac{1}{12}\)

\(\dfrac{2\sqrt2}{1-\sqrt2}-\dfrac{4\sqrt2}{1+\sqrt3}\\ =\dfrac{2\sqrt2(1+\sqrt3)-2\sqrt2(2-2\sqrt2)}{(1-\sqrt2)(1+\sqrt3)}\\ =\dfrac{2\sqrt2(\sqrt3 + 2\sqrt2 -1)}{(1-\sqrt2)(1+\sqrt3)}\)

\(\sqrt{\dfrac{23}{9}}\\ = \dfrac{\sqrt{23}}{\sqrt{9}}\\ =\dfrac{\sqrt{23}}{3}\)

Another approach ---- take log each side.

\(2^x\cdot 5^y = 100 \rightarrow\;\;x\log 2 + y\log 5 = 2\)

\(2^y\cdot 5^x = 1000\rightarrow \;\; y\log 2 + x\log 5 = 3\)

After this we may solve it like a linear equation!!

\(\text{Add the 2 equations up:}\\ x\log 2 + y\log 2 + x\log 5 + y\log 5 = 5\\ (x+y)(\log 2 + \log 5) = 5 \leftarrow\text{Factorization}\\ (x+y)(\log 10) = 5 \leftarrow \text{Property of logarithms}\\ x+y = 5\)

0 seconds. My brain is still working efficiently and I am serious.

Multiply by a negative number.

20, 24, or 28, any of them :D

Thanks Alan and GingerAle :)

Btw what is the correct general formula? :)

\(\boxed{\color{red}\dfrac{1}{a}:\dfrac{1}{b}\color{black}=\color{red}b:a}\)

\(\left(\dfrac{64}{81}\right)^{\frac{-2}{3}}\\ = \left(\dfrac{81}{64}\right)^{\frac{2}{3}}\\ =\dfrac{81^{\frac{2}{3}}}{64^{\frac{2}{3}}}\\ =\dfrac{\sqrt[3]{6561}}{\sqrt[3]{4096}}\\ =\dfrac{\sqrt[3]{6561}}{16}\)