Rationalise the denominator of 1 upon under root 7 +under root 6 -under root 13..
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Plzz frndzzzz tell me the ans by solving it if anybody knows ...it's very much urgent
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1/√7+√6-√13
=1{(√7+√6)+√13}/{(√7+√6)+√13}{(√7+√6)-√13}
=(√7+√6+√13)/{(√7+√6)²-(√13)²}
=(√7+√6+√13)/[{(√7)²+2√7√6+(√6)²}-13]
=(√7+√6+√13)/(7+2√42+6-13)
=(√7+√6+√13)/2√42
=√42(√7+√6+√13)/2√42√42
=√42(√7+√6+√13)/(2×42)
=√42(√7+√6+√13)/84
=1{(√7+√6)+√13}/{(√7+√6)+√13}{(√7+√6)-√13}
=(√7+√6+√13)/{(√7+√6)²-(√13)²}
=(√7+√6+√13)/[{(√7)²+2√7√6+(√6)²}-13]
=(√7+√6+√13)/(7+2√42+6-13)
=(√7+√6+√13)/2√42
=√42(√7+√6+√13)/2√42√42
=√42(√7+√6+√13)/(2×42)
=√42(√7+√6+√13)/84
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