Question

1/(√2+√3)-√4 rarionalizee​

Answer 1

12–√+3–√−4–√  

12–√+3–√−2  

The trick to rationalising with more than 2 numbers in the denominator is to rationalise twice.

Consider the quantity  2–√+3–√  to be one entity.

=1(2–√+3–√)−2⋅(2–√+3–√)+2(2–√+3–√)+2  

Using  a2−b2=(a+b)(a−b)  

=(2–√+3–√)+2(2–√+3–√)2−22  

(2–√+3–√)2=2+3+26–√  

(2–√+3–√)2=5+26–√  

=(2–√+3–√)+25+26–√−22  

=(2–√+3–√)+226–√+1  

Now rationalise this again.

=(2–√+3–√+2)(26–√−1)(26–√)2−12  

=(2–√+3–√+2)(26–√−1)23  

=43–√+62–√+46–√−2–√−3–√−223  

=33–√+52–√+46–√−223  

There.

Answer 2

Answer:

(√2+√3)+√4/1

Step-by-step explanation:

1/(√2+√3)-√4*(√2+√3)+√4/(√2+√3)+√4

(√2+√3)+√4/(√2+√3)^2-(√4)^2

(√2+√3)+√4/(2+3)-4

(√2+√3)+√4/5-4

(√2+√3)+√4/1

or

(√2+√3)+√4