prove that 1 divided by 3 minus root 8 minus 1 / root 8 minus root 7 + 1 divided by root 7 minus root 6 minus 1 divided by root 6 minus root 5 + 1 divided by root 5 minus 2 equals 5
Answers
Given: 1/(3-√8) - 1/(√8 - √7) + 1/(√7 -√6) - 1/(√ 6 - √5) + 1/(√ 5 -2) = 5
To find: Prove the above term is equal to 5.
Solution:
- Lets first consider LHS, we get:
1/(3-√8) - 1/(√8 - √7) + 1/(√7 -√6) - 1/(√ 6 - √5) + 1/(√ 5 -2)
- Lets rationalise all the denominators, we get:
1 x (3+√8) / (3-√8) x (3+√8) - 1 x (√8 + √7) / (√8 - √7) x (√8 +√7)
+ 1 x (√7 +√6) / (√7 -√6) x (√7 +√6) - 1 x (√ 6 + √5) / (√ 6 - √5) x (√ 6 + √5)
+ 1 x (√ 5 +2) / (√ 5 -2) x (√ 5 +2)
- So after rationalising, we get:
(3+√8) / 9-8 - { (√8 + √7)/ 8-7 } + (√7 +√6) / 7-6 - { (√ 6 + √5) / 6-5 } + (√5 +2) / 5-4
- Now solving with denominator as 1, we get:
3+√8- √8 - √7 +√7 +√6 -√ 6 - √5 + √5 +2
- Cancelling the root terms, we get the final answer as:
5
= RHS.
Answer:
So from above solution we have proved that :
1/(3-√8) - 1/(√8 - √7) + 1/(√7 -√6) - 1/(√ 6 - √5) + 1/(√ 5 -2) = 5
Answer:
Given: 1/(3-√8) - 1/(√8 - √7) + 1/(√7 -√6) - 1/(√ 6 - √5) + 1/(√ 5 -2) = 5
To find: Prove the above term is equal to 5.
Solution:
Lets first consider LHS, we get:
1/(3-√8) - 1/(√8 - √7) + 1/(√7 -√6) - 1/(√ 6 - √5) + 1/(√ 5 -2)
Lets rationalise all the denominators, we get:
1 x (3+√8) / (3-√8) x (3+√8) - 1 x (√8 + √7) / (√8 - √7) x (√8 +√7)
+ 1 x (√7 +√6) / (√7 -√6) x (√7 +√6) - 1 x (√ 6 + √5) / (√ 6 - √5) x (√ 6 + √5)
+ 1 x (√ 5 +2) / (√ 5 -2) x (√ 5 +2)
So after rationalising, we get:
(3+√8) / 9-8 - { (√8 + √7)/ 8-7 } + (√7 +√6) / 7-6 - { (√ 6 + √5) / 6-5 } + (√5 +2) / 5-4
Now solving with denominator as 1, we get:
3+√8- √8 - √7 +√7 +√6 -√ 6 - √5 + √5 +2
Cancelling the root terms, we get the final answer as:
5
= RHS.
Step-by-step explanation: