Prove that- i a-b-c-a-1b-1-b-1c-1-c-1a-1-abc ii a-1-b-1-1-ab-a-b

(i) a+b+c/a^ 1b^ 1+b^ 1c^ 1+c^ 1a^ 1=abc LHS=a+b+c/a^ 1b^ 1+b^ 1c^ 1+c^ 1a^ 1 ∵ a^ m=1/a^m =a+b+c/1/ab+1/bc+1/ca=a+b+c/c+a+b/abc =(a+b+c)abc/(a+b+c)=abc ...

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Byju's Answer

Standard IX

Mathematics

Laws of Exponents for Real Numbers

Prove that: i...

Question

Prove that:

(i) $\frac{a + b + c}{a^{- 1} b^{- 1} + b^{- 1} c^{- 1} + c^{- 1} a^{- 1}} = a b c$

(ii) $(a^{- 1} + b^{- 1})^{- 1} = \frac{a b}{a + b}$

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Solution

(i) $\frac{a + b + c}{a^{- 1} b^{- 1} + b^{- 1} c^{- 1} + c^{- 1} a^{- 1}} = a b c$

$L H S = \frac{a + b + c}{a^{- 1} b^{- 1} + b^{- 1} c^{- 1} + c^{- 1} a^{- 1}} {\begin{matrix} ∵ a^{- m} = \frac{1}{a^{m}} \end{matrix}} = \frac{a + b + c}{\frac{1}{a b} + \frac{1}{b c} + \frac{1}{c a}} = \frac{a + b + c}{\frac{c + a + b}{a b c}} = \frac{(a + b + c) a b c}{(a + b + c)} = a b c = R H S$

(ii) $(a^{- 1} + b^{- 1})^{- 1} = \frac{a b}{a + b}$

$L H S = (a^{- 1} + b^{- 1})^{- 1} = {(\frac{1}{a} + \frac{1}{b})}^{- 1} = {(\frac{b + a}{a b})}^{- 1} {\begin{matrix} ∵ a^{- m} = \frac{1}{a^{m}} \end{matrix}} = \frac{a b}{a + b} = R H S$

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Similar questions

Q. Prove that:

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\frac{a + b + c}{a^{- 1} b^{- 1} + b^{- 1} c^{- 1} + c^{- 1} a^{- 1}} = a b c

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Q. Let a, b, c be such that b

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∣ ∣ ∣ ∣ \begin{matrix} a & a + 1 & a - 1 - b & b + 1 & b - 1 c & c - 1 & c + 1 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} a + 1 & b + 1 & c - 1 a - 1 & b - 1 & c + 1 (- 1)^{n + 2} a & (- 1)^{n + 1} b & (- 1)^{n} c \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

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Prove that: (i) a+b+ca−1b−1+b−1c−1+c−1a−1=abc (ii) (a−1+b−1)−1=aba+b

(i) a+b+ca−1b−1+b−1c−1+c−1a−1=abc LHS=a+b+ca−1b−1+b−1c−1+c−1a−1 {∵a−m=1am}=a+b+c1ab+1bc+1ca=a+b+cc+a+babc=(a+b+c)abc(a+b+c)=abc=RHS (ii) (a−1+b−1)−1=aba+b LHS=(a−1+b−1)−1=(1a+1b)−1=(b+aab)−1 {∵a−m=1am}=aba+b=RHS

Prove that:

(i) $\frac{a + b + c}{a^{- 1} b^{- 1} + b^{- 1} c^{- 1} + c^{- 1} a^{- 1}} = a b c$

(ii) $(a^{- 1} + b^{- 1})^{- 1} = \frac{a b}{a + b}$