Question

Simplify 1 /√7−√6 — 1 /√6−√5 + 1 /√5−2 + 1 /2−√3 − 1 /3−√2
First correct answer i will make brainliest
please do fast

Answer 1

Answer:

1+√3+√5)⋅(1+√3−√5) . We can write this as ((1+√3)−√5)(1+√3)+√5) . This can then be made is DOPS and written as : (1+√3)2−(√5)2 . Simplify that to ... 11+√3−√5=7+3√3+√5+2√1511 ...

Step-by-step explanation:

Answer 2

Step-by-step explanation:

$\begin{gathered} \tt \: \frac{1}{1 + {x}^{b - a} + {x}^{c - a} } + \frac{1}{1 + {x}^{a - b} + {x}^{c - b} } + \frac{1}{1 + {x}^{b - c} {x}^{a - c} } \\ \\ \\ \\ : \implies \: \tt \: \frac{ {x}^{a} }{ {x}^{a} \bigg(1 + {x}^{b - a} + {x}^{c - a} \bigg) } + \frac{ {x}^{b} }{ {x}^{b} \bigg(1 + {x}^{a - b} + {x}^{c - b} \bigg)} + \frac{ {x}^{c} }{ {x}^{c} \bigg( 1 + {x}^{b - c} + {x}^{a - c} \bigg) } \\ \\ \\ \\ : \implies \tt \frac{ {x}^{a} }{ {x}^{a} + {x}^{b} + {x}^{c} } + \frac{ {x}^{b} }{ {x}^{b} + {x}^{a} + {x}^{c} } + \frac{ {x}^{c} }{ {x}^{c} + {x}^{b} + {x}^{a} } \\ \\ \\ \\ : \implies \tt \bigg(\frac{ \cancel{{x}^{a} + {x}^{b} + {x}^{c}} }{ \cancel{ {x}^{a} + {x}^{b} + {x}^{c} } } \bigg) \\ \\ \\ \\ \bf \: \large : \implies \large \: {\tt 1 \: \: \: \: \bf \bigg [\:Proved\: \bigg]}\end{gathered}$