Question

Pls Answer this question for me pls​

Answer 1

Step-by-step explanation:

1/(3+√7)

on rationlizing

1/(3+√7)×(3-√7)/(3-√7)

(3-√7)/2

1/(√7+√5)

on rationlizing

1/(√7+√5)×(√7-√5)/(√7+√5)

(√7-√5)/2

1/(√5+√3)

on rationlizing

1/(√5+√3)×(√5-√3)/(√5-√3)

(√5-√3)/2

1/(√3+1)

on rationlizing

1/(√3+1)×(√3-1)/(√3-1)

(√3-1)/2

now,

1/(3+√7)+1/(√7+√5)+1/(√5+√3)+1/(√3+1)=a+b√5

(3-√7)/2 +(√7-√5)/2 +(√5-√3)/2 +(√3-1)/2=a+b√5

(3-√7+√7-√5+√5-√3+√3-1)/2= a+b√5

2/2= a+b√5

1= a+b√5

1+0= a+b√5

hence a=1 and b=0

#666