here's the problem.
i don't know how to solve this , can you help me? thanks.
$\displaystyle \frac{(^4\sqrt{4b^4})+(^4\sqrt{9b^4}-(\sqrt{12a^2}-(^4\sqrt{64a^4}+(\sqrt{20a^2}-(\sqrt{5b^2})}{b-2a}^{-1}$
here's the problem.
i don't know how to solve this , can you help me? thanks.
$\displaystyle \frac{(^4\sqrt{4b^4})+(^4\sqrt{9b^4}-(\sqrt{12a^2}-(^4\sqrt{64a^4}+(\sqrt{20a^2}-(\sqrt{5b^2})}{b-2a}^{-1}$
The first exemple must be an olympic exercise:
$\displaystyle \left(\dfrac{\sqrt[4]{4b^4} + \sqrt[4]{9b^4}-\sqrt{12a^2}-\sqrt[4]{64a^4} + \sqrt{20a^2} - \sqrt{5b^2}}{b-2a} \right)^{-1} = $
$\displaystyle \left(\dfrac{b\sqrt{2} + b\sqrt{3}-2a\sqrt{3}-2a\sqrt{2} + 2a\sqrt{5} - b\sqrt{5}}{b-2a} \right)^{-1} = $
$\displaystyle \left(\dfrac{b \left(\sqrt{2} + \sqrt{3}-\sqrt{5}\right)-2a \left(\sqrt{3}+\sqrt{2} - \sqrt{5}\right) }{b-2a} \right)^{-1} = $
$\displaystyle \left(\dfrac{ \left(\sqrt{2} + \sqrt{3}-\sqrt{5}\right)(b-2a) }{b-2a} \right)^{-1} = \dfrac1{\sqrt{2} + \sqrt{3}-\sqrt{5}}$
Now eliminate the square roots in the denominator:
$\displaystyle \dfrac1{\sqrt{2} + \sqrt{3}-\sqrt{5}} \cdot \dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{\sqrt{2} + \sqrt{3}+\sqrt{5}} = \dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{2+2\sqrt{6}+3-5} = $
$\displaystyle \dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{2\sqrt{6}} = \dfrac1{12} \cdot \left(2\sqrt{3}+3\sqrt{2} + \sqrt{30} \right)$
phooo!
The next exemple is much easier. Show your work and I'M going to help you if you don't succeed. (You should get $\displaystyle 2\sqrt{6}$)
I'll show you how to do the second question, but if I shall help you I must know what exactly is confusing you.
Ooooh, I finally found your question in the quoted calculations.
If you have $\displaystyle \left(\dfrac{\sqrt[4]{4b^4} + \sqrt[4]{9b^4}-\sqrt{12a^2}-\sqrt[4]{64a^4} + \sqrt{20a^2} - \sqrt{5b^2}}{b-2a} \right)^{-1} = $ then you can simplify
$\displaystyle \sqrt[4]{4b^4} = (2^2 \cdot b^4)^{\frac14} = 2^{\frac24} \cdot b^{\frac44} = b\cdot 2^{\frac12} = b\cdot \sqrt{2}$
Apply similar transformations with all summands in the numerator and you'll get the next line of my calculations.
And now the 2nd question:
$\displaystyle \left(\dfrac{a}{\sqrt{2}} \div (2a\sqrt{3})\right)^{-1} = \left(\dfrac{a}{\sqrt{2}} \cdot \dfrac1{2a\sqrt{3}}\right)^{-1} =$ $\displaystyle \left(\dfrac1{2\sqrt{6}} \right)^{-1} = 2\sqrt{6}$
Unfortunately you quoted me wrong:
$\displaystyle
\dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{2\sqrt{6}} = \dfrac{{\color{red}\bold{\sqrt{6}}} (\sqrt{2} + \sqrt{3}+\sqrt{5})}{2\sqrt{6} \cdot {\color{red}\bold{\sqrt{6}}}} = \dfrac{\sqrt{12}+\sqrt{18} + \sqrt{30}}{12} =$ $\displaystyle \dfrac1{12} \cdot \left(2\sqrt{3}+3\sqrt{2} + \sqrt{30} \right)
$
yeah I got it thanks
I'll try to continue from here
$\displaystyle \frac{2\sqrt{3}+3\sqrt{2}+\sqrt{30}}{12} * (2\sqrt{6})$
$\displaystyle \frac{4\sqrt{18}+6\sqrt{12}+2\sqrt{180}}{12}$
$\displaystyle \frac{12\sqrt{2}+12\sqrt{3}+12\sqrt{5}}{12}$
is this correct?
Honestly, I don't know what you did and why?
All my calculations were necessary to get a rational denominator. Thus I multiplied numerator and denominator by $\displaystyle \sqrt{6}$ which I marked in red. That means I didn't change the value of the quotient.
$\displaystyle 2\cdot \sqrt{6}\cdot \sqrt{6} = 12$
The product of the numerator and $\displaystyle \sqrt{6}$ is
$\displaystyle \sqrt{12}+\sqrt{18} + \sqrt{30} = \sqrt{4 \cdot 3}+\sqrt{9 \cdot 2} + \sqrt{30}$ which I simplified to
$\displaystyle 2\sqrt{3}+3\sqrt{2} + \sqrt{30}$
That's all.