Question

3-√3 делённое на √6-√2 сколько будет?

Answer 1

$\frac{3- \sqrt{3} }{ \sqrt{6} - \sqrt{2} } = \frac{(3- \sqrt{3})( \sqrt{6} + \sqrt{2}) }{( \sqrt{6} - \sqrt{2})( \sqrt{6} + \sqrt{2}) }= \frac{3 \sqrt{6}+3 \sqrt{2} - \sqrt{18}- \sqrt{6} }{ 6-2} = \frac{2 \sqrt{6}+3 \sqrt{2} - 3\sqrt{2}}{4} = \\\\ = \frac{2 \sqrt{6} }{4}= \frac{ \sqrt{6} }{2}$

Answer 2

To simplify the expression (3-√3) / (√6-√2), we can use the rationalizing technique. This involves multiplying both the numerator and the denominator by the conjugate of the denominator, which is (√6+√2).

So, the expression becomes:

(3-√3) / (√6-√2) * (√6+√2) / (√6+√2)

Now, we can use the distributive property to multiply the numerators and denominators:

= (3√6 + 3√2 - √18 - √6) / (6 - 2)

Now, simplify the expression further:

= (2√6 + 3√2 - √18) / 4

Next, we can simplify the radical:

√18 = √(9 * 2) = 3√2

Substitute this back into the expression:

= (2√6 + 3√2 - 3√2) / 4

The √2 terms in the numerator cancel out:

= 2√6 / 4

Finally, simplify:

= √6 / 2

So, (3-√3) / (√6-√2) is equal to √6 / 2.