यदि x² + y² + z² = xy + yz + zx और x = 1 है, तो \(\rm \frac{10x^4+5y^4+7z^4}{13x^2y^2+6y^2z^2+3z^2x^2}\) का मान ज्ञात कीजिए?

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Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 05 Dec 2022 Shift 4)

Answer 1

दिया गया:

x = 1

x2 + y2 + z2 = xy + yz + zx

गणना:

x2 + y2 + z2 - xy - yz - zx = 0

⇒(1/2)[(x - y)2 + (y - z)2 + (z - x)2] = 0

⇒x = y , y = z और z = x

लेकिन x = y = z = 1

इसलिए, \(\rm \frac{10x^4+5y^4+7z^4}{13x^2y^2+6y^2z^2+3z^2x^2}\)

= {10(1)4 + 5(1)4 + 7(1)4}/{13(1)2(1)2+ 6(1)2(1)2 + 3(1)2(1)2}

= 22/22

= 1

इसलिए, अभीष्ट मान 1 है।

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