जर abc = 5 असेल, तर (\({1 \over 1 \ + \ a \ + \ b^{-1} }\) + \({1 \over 1 \ + \ b \ + \ 5 c^{-1}}\) +\(\frac{1}{1+\frac{c}{5}+a^{-1}}\) \(\)) चे मूल्य काय आहे?

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Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 08 Dec 2022 Shift 3)

Answer 1

दिलेल्याप्रमाणे:

जर abc = 5

वापरलेली संकल्पना:

abc = 5

⇒ ac = \(5\over b \)

⇒ (b)^-1

⇒ b = \(5\over ac\)

गणना:

\({1 \over 1 \ + \ a \ + \ b^{-1} }\) + \({1 \over 1 \ + \ b \ + \ 5 c^{-1}}\) + \(\frac{1}{1+\frac{c}{5}+a^{-1}}\)

⇒\(\frac{1}{1+\frac{ac}{5}+a}\) + \(\frac{1}{1+\frac{5}{ac}+5c^{-1}}\) + \(\frac{1}{1+\frac{1}{ab}+a^{-1}}\)

⇒\(5\over5+5a+ac \)\(+\) \(ac\over5+5a+ac \) \(+\) \(a\over5+a+ac \)

⇒\(a+5+ac\over5+a+ac \)

⇒\(1\)

म्हणून, ( \({1 \over 1 \ + \ a \ + \ b^{-1} }\)+\({1 \over 1 \ + \ b \ + \ 5 c^{-1}}\)+\(\frac{1}{1+\frac{c}{5}+a^{-1}}\)) चे मूल्य 1 आहे.

Shortcut Trick a = b = 1 आणि c = 5 समजा

आता, प्रश्नानुसार,

⇒ \({1 \over 1 \ + \ 1 \ + \ 1}\) + \({1 \over 1 \ + \ 1 \ + {5 \over 5 }}\) + \(\frac{1}{1+\frac{5}{5}+ {1}}\)

⇒ 1/3 + 1/3 + 1/3

⇒ 3/3 = 1