if $x^2+y^2 =z^2 =xy+yz+zx $ and x

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Target Exam

CUET

Subject

General Test

Chapter

Quantitative Reasoning

Topic

Algebra

Question:

if $x^2+y^2 =z^2 =xy+yz+zx$ and x = 1, then find the value of $\frac{10x^4+5y^4+7z^4}{13x^2y^2+6y^2z^2+3z^2x^2}$ .

Options:

-1

Correct Answer:

Explanation:

if $x^2+y^2 =z^2 =xy+yz+zx$

Then put the value of x = y = z = 1

$\frac{10x^4+5y^4+7z^4}{13x^2y^2+6y^2z^2+3z^2x^2}$ = $\frac{10×1^4+5×1^4+7×1^4}{13×1^2×1^2+6×1^2×1^2+3×1^2×1^2}$

= $\frac{22}{22}$ = 1