Asked by tanvir
remove the irrationality in the denominator
(i) 1/1+√2+√3
(i) 1/1+√2+√3
Answers
Answered by
Steve
1/(1+√2+√3) * 1/(1-(√2+√3))/(1-(√2+√3))
= (1-√2-√3)/(1 - (√2+√3)^2)
= (1-√2-√3)/(-4-2√6)
= (1-√2-√3)/(-2(2+√6))
= (1-√2-√3)/(-2(2+√6)) * (2-√6)/(2-√6)
= (1-√2-√3)(2-√6)/-2 * 1/(4-6)
= (2+√2-√6)/4
= (1-√2-√3)/(1 - (√2+√3)^2)
= (1-√2-√3)/(-4-2√6)
= (1-√2-√3)/(-2(2+√6))
= (1-√2-√3)/(-2(2+√6)) * (2-√6)/(2-√6)
= (1-√2-√3)(2-√6)/-2 * 1/(4-6)
= (2+√2-√6)/4
Answered by
Priyanshu
Right
Answered by
HRITIK
Right
Answered by
Hrishi
Good
Answered by
Hrishi
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