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We have $$\dfrac{4^{0.75}}{1+\sqrt2+\sqrt3}+9^{0.25}=\dfrac{2\sqrt2}{1+\sqrt2+\sqrt3}+\sqrt3$$ To evaluate this, the standard approach is multiplying the fraction by $\dfrac{1-\sqrt2-\sqrt3}{1-\sqrt2-\sqrt3}$ and then we left with one square root in the denominator and by multiplying the fraction by the conjugate of the denominator again we get rid of all square roots in denominator. but my question is : Is there another approach (preferably faster one) to evaluate the expression?

I also tried using the common denominator and get this:

$$\dfrac{2\sqrt2+\sqrt3+\sqrt6+3}{1+\sqrt2+\sqrt3}$$ It seems I can't proceed from here without using the previous method.

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  • $\begingroup$ No................... $\endgroup$ Mar 30, 2021 at 18:20
  • $\begingroup$ It's not clear what kind of answer you are looking for. $\endgroup$
    – TonyK
    Mar 30, 2021 at 18:35
  • $\begingroup$ use FullSimplify[2 Sqrt[2]/(1 + Sqrt[2] + Sqrt[3]) + Sqrt[3]] in Mathematica to get 1+Sqrt[2]. $\endgroup$
    – Somos
    Mar 30, 2021 at 18:44
  • $\begingroup$ @Somos How is that possible. There is no $\sqrt{3}$ at the end? $\endgroup$
    – nonuser
    Mar 30, 2021 at 18:48

2 Answers 2

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Start with $$ \frac{2\sqrt2}{1+\sqrt2+\sqrt3}+\sqrt3. \tag{1 }$$ Put them both in common denominator $$ \frac{2\sqrt2}{1+\sqrt2+\sqrt3} + \frac{\sqrt3(1+\sqrt2+\sqrt3)}{1+\sqrt2+\sqrt3}. \tag{2} $$ Add the two fractions $$ \frac{2\sqrt2 + (\sqrt3+\sqrt6+3)}{1+\sqrt2+\sqrt3}.\tag{3} $$ Collect the numerator $$ \frac{3 + 2\sqrt2 + \sqrt3+\sqrt6}{1+\sqrt2+\sqrt3}.\tag{4} $$ Take out $1$ from the fraction $$ 1 + \frac{\sqrt2 + 2 + \sqrt6}{1+\sqrt2+\sqrt3}. \tag{5}$$ Now recognize that the numerator is divisible by the denominator. $$ 1 + \sqrt2. \tag{6}$$

In general, using the same reasoning we get $$ \frac{(2\!+\!a\!-\!n)\!+\!(a\!+\!1)\sqrt2 \!+\!(a\!-\!1)\sqrt{n}} {1+\sqrt2+\sqrt{n}}\!+\!\sqrt{n} \!=\! a \!+\! \sqrt2. $$ Make the substitutions $\,a=1,n=3\,$ to get our special case.

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Since $$4^{0.75}=4^\frac34=\left(2^2\right)^\frac34=2^\frac32=\sqrt{2^3}=2\sqrt2$$

and

$$9^{0.25}=9^\frac14=\left(3^2\right)^\frac14=3^\frac12=\sqrt3\;,$$

it follows that

$$\dfrac{4^{0.75}}{1+\sqrt2+\sqrt3}+9^{0.25}=\dfrac{2\sqrt2}{1+\sqrt2+\sqrt3}+\sqrt3\;.$$

Moreover,

$$\dfrac{2\sqrt2}{1+\sqrt2+\sqrt3}+\sqrt3=\dfrac{\left(3+2\sqrt2\right)-3}{1+\sqrt2+\sqrt3}+\sqrt3=$$

$$=\dfrac{\left(1+\sqrt2\right)^2-\left(\sqrt3\right)^2}{1+\sqrt2+\sqrt3}+\sqrt3=$$

$$=\dfrac{\left(1+\sqrt2+\sqrt3\right)\left(1+\sqrt2-\sqrt3\right)}{1+\sqrt2+\sqrt3}+\sqrt3=$$

$$=\left(1+\sqrt2-\sqrt3\right)+\sqrt3=1+\sqrt2\;.$$

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    $\begingroup$ You are just repeating what the OP already said. $\endgroup$
    – TonyK
    Mar 30, 2021 at 18:38
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    $\begingroup$ Thanks, But I already knew this. my question was how to calculate $\dfrac{2\sqrt2}{1+\sqrt2+\sqrt3}+\sqrt3$ without using conjugates method. $\endgroup$
    – Amirali
    Mar 30, 2021 at 18:38
  • $\begingroup$ I have calculated your expression without using conjugates method. $\endgroup$
    – Angelo
    Mar 30, 2021 at 19:12
  • $\begingroup$ Why have you downvoted my answer? Is there any mistakes? If yes, what are my mistakes? $\endgroup$
    – Angelo
    Mar 30, 2021 at 19:22
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    $\begingroup$ I explained why I downvoted your answer. Since then, you have edited it. $\endgroup$
    – TonyK
    Mar 30, 2021 at 19:25

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