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if $x^2+y^2 =z^2 =xy+yz+zx $ and x = 1, then find the value of $\frac{10x^4+5y^4+7z^4}{13x^2y^2+6y^2z^2+3z^2x^2}$. |
2 0 -1 1 |
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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 |
