simplify: 1/(1+√2+√3)+(1/(1-√2+√3)
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19
(1/1+√2+√3) + (1/1-√2+√3)
=(1-√2+√3+1+√2+√3)/(1+√2+√3)(1-√2+√3)
=(2+2√3)/(1-√2+√3+√2-2+√6+√3-√6+3)
=(2+2√3)/(2+2√3)
=1
=(1-√2+√3+1+√2+√3)/(1+√2+√3)(1-√2+√3)
=(2+2√3)/(1-√2+√3+√2-2+√6+√3-√6+3)
=(2+2√3)/(2+2√3)
=1
Aldrago:
pls solve my next question
Answered by
4
1/(1+√2+√3) + 1/(1-√2+√3)
=1+√2+√3+1-√2+√3/(1+√3+√2)(1+√3-√2)
=2+2√3/(1+√3)^2×(√2)^2['.'(a+b)(a-b)=(a^2+b^2)]
=2(1+√3)/(1+2√3+3)2
=1+√3/1+2√3+3
=1+√2+√3+1-√2+√3/(1+√3+√2)(1+√3-√2)
=2+2√3/(1+√3)^2×(√2)^2['.'(a+b)(a-b)=(a^2+b^2)]
=2(1+√3)/(1+2√3+3)2
=1+√3/1+2√3+3
its wrong
sorry
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