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# 6^(1/2)/(1/2^(1/2)+1/3^(1/2))

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AbdurRakib
$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$

A. $$\frac{6}{5}$$

B. $$\frac{5}{6}$$

C. 5

D. $$2\sqrt{3} +3\sqrt{2}$$

E. $$6\sqrt{3}-6\sqrt{2}$$

Nice question!

Here's another approach:

Let's first deal with the denominator: 1/√2 + 1/√3 = √3/√6 + √2/√6
= (√3 + √2)/√6

So....$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$ √6/[(√3 + √2)/√6]

= (√6)(√6)/(√3 + √2)

= 6/(√3 + √2)

Now we can use the fact that √3 ≈ 1.7 and √2 ≈ 1.4 to get....
≈ 6/(1.7 + 1.4)
≈ 6/(3.1)

Since 6/3 = 2, we know that 6/(3.1) = a number a bit smaller than 2

Now check the answer choices to see which one evaluates to be a number a bit smaller than 2

A. $$\frac{6}{5}$$ ELIMINATE

B. $$\frac{5}{6}$$ ELIMINATE

C. 5 ELIMINATE

D. $$2\sqrt{3} +3\sqrt{2}$$ ≈ 2(1.7) + 3(1.4) ≈ a number that's bigger than 6 ELIMINATE

E. $$6\sqrt{3}-6\sqrt{2}$$[/quote] = 6(√3 - √2) ≈ 6(1.7 - 1.4) ≈ 6(0.3) ≈ 1.8 PERFECT!

Cheers,
Brent
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AbdurRakib
$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$

A. $$\frac{6}{5}$$

B. $$\frac{5}{6}$$

C. 5

D. $$2\sqrt{3} +3\sqrt{2}$$

E. $$6\sqrt{3}-6\sqrt{2}$$

$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$

So if you Multiply the ratio with $$\sqrt{3}-\sqrt{2}$$

You will get the ratio as 6 ($$\sqrt{3}-\sqrt{2}$$) / $$\sqrt{3}-\sqrt{2}$$ * $$\sqrt{3}+\sqrt{2}$$

6 * ($$\sqrt{3}-\sqrt{2}$$ ) / 3-2

Using $$a^2 - b^2 = (a-b) (a+b)$$

6 * ($$\sqrt{3}-\sqrt{2}$$)

E
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