\(\rm \frac{1}{x(x-y)(x-z)}+\frac{1}{y(y-z)(y-x)}+\frac{1}{z(z-x)(z-y)}\) எதற்கு சமம் ?

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CDS 02/2022 Elementary Mathematics Official Paper (Held On 04 Sep 2022)
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  1. 0
  2. 1
  3. \(\rm\frac{1}{xyz}\)
  4. \(-\rm\frac{1}{xyz}\)

Answer (Detailed Solution Below)

Option 3 : \(\rm\frac{1}{xyz}\)
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நம்மிடம் உள்ளது,

\(\rm \frac{1}{x(x-y)(x-z)}+\frac{1}{y(y-z)(y-x)}+\frac{1}{z(z-x)(z-y)}\)

x = 1, y = 2, z = 3 ஐ பிரதியிட 

\(\rm \frac{1}{1(1-2)(1-3)}+\frac{1}{2(2-3)(2-1)}+\frac{1}{3(3-1)(3-2)}\)

\(\frac{1}{(-1)(-2)}+\frac{1}{2(-1)}+\frac{1}{3(2)}\)

\(\frac{1}{2}-\frac{1}{2}+\frac{1}{6} = \frac{1}{6}\)

விருப்பம் (3) இல் இருந்து

\(\frac{1}{xyz} = \frac{1}{1\times 2\times 3} = \frac{1}{6}\)

\(\rm\frac{1}{xyz}\) என்பது சரியான மதிப்பு.

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