Math question HELP?

Question

Rationalize the denominator
one over the squre root of 2 plus one
show me how you work this

Answer 1

1/(√2+1)
Multiply by (√2-1)/(√2-1):
(√2-1)/((√2+1)(√2-1))
Expand:
(√2-1)/(2+√2-√2-1)
Simplify:
(√2-1)/(1)
√2-1

Answer 2

i don't know if you mean a) 1 / (squareroot of 2 plus 1) or
b) (1/squareroot of 2) + 1.

anyway, in the first case:

1/(sqrt2 + 1) ===> you multiply the numerator and denominator by its conjugate: (sqrt2 - 1).
thus:

1 times (sqrt2 - 1) ===> sqrt2 -1 over
(sqrt2 +1) times (sqrt2 -1) ===> 2 - 1 or simply 1
recall: (x+y)(x-y) = x^2 - y^2

thus, the answer is sqrt2 - 1.

for case b:

(1/sqrt2) + 1 this is easier.
first, combine the fraction and 1 with sqrt2 as the common denominator.

(1/sqrt2) + 1 ===> 1 + sqrt2 over sqrt2 or (1+sqrt2)/sqrt2
to rationalize, multiply the numerator and denominator by sqrt2:

(1+sqrt2)(sqrt2) over 2.
or:
(sqrt2 + 1)/2



note: the secret of the first solution is the multiplication of its conjugate.

Answer 3

Rationalise 1/(sqrt(2)+1) I assume

Note sqrt(2) x sqrt(2) = 2 Obviously and

(x - y)(x + y) = x^2 - y^2

a x b/b = a x 1 = a

Thusly

1/(sqrt(2)+1) x (sqrt(2)-1)/(sqrt(2)-1)

=(sqrt(2)-1)/(2 -1)

=sqrt(2) - 1

Follow the steps. Easy peasy.

Onya

Answer 4

for rationalising you multiply the denominator by the conjugate,in this case rt2-1 and in order that you don't change the value of the expression you multiply the numerator also by rt2-1
(rt2+1)(rt2-1)=[(rt2)^2-(1)^2] =2-1=1
[using the identity (a+b)(a-b)=(a^2-b^2).]
so the answer=(rt2-1)

Answer 5

1/(sqrt(2) +1)

multiply by (sqrt(2) - 1)/(sqrt(2) -1) which is supposed to be 1
(sqrt(2) -1)/ ((sqrt(2) + 1)(sqrt(2) -1))

= sqrt(2) - 1


idea is that (a+1)(a-1) = (a^2 -1)
so if a is sqrt(2), a^2 = 1 so a^2 - 1 = 1

Answer 6

1/ (sq rt(2) + 1)
multiply this by (sq rt(2) - 1) / (sq rt(2) - 1)
this should get rid of the radical symbol in the denominator
work it out:
(sq rt(2) - 1) / (2 - 1)

Answer:
(sq rt(2) - 1)