Math, asked by prasannatanish, 9 months ago

a+b+c/(a^-1×b^-1)+(b^-1×c^-1)+(c^-1×a^-1)=abc​

Answers

Answered by Uditdon6666
1

Step-by-step explanation:

a+b+c/(a^-1×b^-1)+(b^-1×c^-1)+(c^-1×a^-1)=abca+b+c/(a^-1×b^-1)+(b^-1×c^-1)+(c^-1×a^-1)=abca+b+c/(a^-1×b^-1)+(b^-1×c^-1)+(c^-1×a^-1)=abca+b+c/(a^-1×b^-1)+(b^-1×c^-1)+(c^-1×a^-1)=abca+b+c/(a^-1×b^-1)+(b^-1×c^-1)+(c^-1×a^-1)=abca+b+c/(a^-1×b^-1)+(b^-1×c^-1)+(c^-1×a^-1)=abc

Answered by rekhadevi5392
5

Answer:

To Prove:a+b+c/(a^-1×b^-1)+(b^-1×c^-1)+(c^-1×a^-1)=abc

Proof:

a+b+c/(a^-1×b^-1)+(b^-1×c^-1)+(c^-1×a^-1)

a+b+c/[(ab)^-1]+[(bc)^-1]+[(ca)^-1]

a+b+c/(1/ab)+(1/bc)+(1/ca)

a+b+c/(c+a+b/abc)

a+b+c/(a+b+c/abc)

a+b+c÷(a+b+c/abc)

a+b+c×abc/a+b+c

abc.

LHS = RHS

proved

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