Can someone help me with this square root equation?

Question

(√3 + 1) ÷ (√2 +1)

(It reads: The square root of three plus one divided by the square root of two plus one.)

Thanks.

Answer 1

By Rationalizing you get

√6 + √2 - √3 - 1

Answer 2

Whenever you have something like a + √b in the denominator, you can always make it rational by multiplying by the conjugate, a - √b.  So multiply the top and bottom here by √2 - 1.

[(√3 + 1) / (√2 + 1)] * (√2 - 1)/(√2 - 1) =

[√6 + √2 + √3 + 1 ] / (2 - 1)

√6 + √2 + √3 + 1

Answer 3

Multiply top and bottom by √2 - 1 to make the denominator rational.

(√3 + 1)(√2 - 1)/((√2 + 1)(√2 - 1))

= (√6 +√2 - √3 + 1) / (4 - 1)

= 1/3(√6 +√2 - √3 + 1)

I'm not sure that this is any better than the original :-)

--- EDITED to add ---

I've seen quite a few answers below this that seem to be quite emphatic about rationalizing the denominator, but having done so they throw it away!  The denominator value becomes 3, so you need a 1/3 factor in your final answer.

Answer 4

(√3 + 1) / (√2 +1)

rationalising

(√3 + 1) (- √2 +1) / (√2 +1)(- √2 +1)

(-√6 - √2 + √3 + 1) / (1 - 2)

(√6 + √2 - √3 - 1)

(√6 - √3 + √2  - 1)

√3(√2 - 1) + 1 (√2 - 1)

(√3 +1) (√2 - 1)

Answer 5

I'm guessing you want me to "rationalize" this expression.

1) Multiply top and bottom of the fraction by the denominators conjugate, (1 - √2).

2) After doing step 1, you should be left with -(1 + √3)(1 - √2)

Answer 6

(√3 + 1)(√2 - 1)/(√2 +1)(√2 -1)

(√6-√3+√2 -1)/(2-1)

(√6-√3+√2 -1)

√3(√2 - 1) + 1 (√2 - 1)

(√3 +1) (√2 - 1)

Answer 7

(√3 + 1)  = 2.732, rounded

(√2 +1)  = 2.414, rounded

(√3 + 1) ÷ (√2 +1) = 1.132, rounded

Answer 8

√3 + 1.....√2 - 1

---------- * -----------

√2 + 1.....√2 - 1

= √6 + √2 - √3 - 1