Question

how to rationalize 1/√7+√3-√2​

Answer 1

$\frac{1}{ \sqrt{7} + \sqrt{3} - \sqrt{1} }$

to rationalise the denominator of tri terms...

we have to change the signs...

therefore...

$\frac{1}{ \sqrt{7} + \sqrt{3} - \sqrt{1} } \times \frac{1}{ \sqrt{7} - \sqrt{3} + \sqrt{1} }$

then,

$\frac{1}{ { \sqrt{7} }^{2} + { \sqrt{3} }^{2} - { \sqrt{1} }^{2} }$

then,

$\frac{1}{7 + 3 + 1}$

then,

$\frac{1}{9}$

Its over.

We have rationalized the denominator..

Answer 2

Step-by-step explanation:

How do you rationalise 1/root7+root3-root2?

I'm going to go with the assumption that the intent of the person asking was how to rationalize the denominator in the following expression:

17√+3√−2√

If the original intent was for

17√+3–√−2–√

then Ms. Stapel has a better answer than I could make.

First, multiply by 7√+3√+2√7√+3√+2√

The result is

7√+3√+2√(7√+3√−2√)(7√+3√+2√)=7√+3√+2√(7√+3√)2−2

Expanding the portion in parentheses, you get

7√+3√+2√7+221√+3−2=7√+3√+2√8+221√

The next step is to multiply by 8−221√8−221√

The result is

(7√+3√+2√)(8−221√)(8+221√)(8−221√)=(7√+3√+2√)(8−221√)64−(4)(21)=(7√+3√+2√)(8−221√)−20

You can factor out a 2 from within the second set of parenthesis of the numerator, and cancel that 2 out with the denominator.

(7√+3√+2√)(4−21√)−10

Multiply the parenthesis.

47√−7√21√+43√−3√21√+42√−2√21√−10

Multiply products involving 2 radicals.

47√−147√+43√−63√+42√−42√−10

Simplify radicals that have perfect square factors.

47√−73√+43√−37√+42√−42√−10

Collect terms in the numerator.

7√−33√+42√−42√−10

Rearrange the numerator to put the radicals in numerical order from the largest to the smallest.

−42√+7√−33√+42√−10

Multiply by −1−1.

42√−7√+33√−42√10