If $x+y+z=0, \quad \frac{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}{x^{4}+y^{4}+z^{4}}=?$

$\frac{1}{2}$

$\frac{3}{2}$

Answer : Option C

Explanation :

x + y + z = 0

\Rightarrow(x+y+z)^{2}=0

\Rightarrow x^{2}+y^{2}+z^{2}+2(x y+y z+z x)=0

\Rightarrow x^{2}+y^{2}+z^{2}=-2(x y+y z+z x)

दोनों पक्षों का वर्ग करने पर, (x^{^{2 + y^{^{2 + z^{^{2)^{^{2 = 4 (xy + yz + zx)^{^{2 $\left(x^{2}+y^{2}+z^{2}\right)^{2}=4(x y+y z+z x)^{2}$ $=4\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+2 x y^{2} z+2 x y z^{2}+2 x^{2} y z\right)$ $\begin{aligned} \Rightarrow & x^{4}+y^{4}+z^{4}=2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+2 x y z(x+y\\ &+z) \end{aligned}$ $\begin{aligned} \Rightarrow & x^{4}+y^{4}+z^{4}=2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right) \\ &[\because x+y+z=0] \end{aligned}$ $\Rightarrow \frac{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}{x^{4}+y^{4}+z^{4}}=\frac{1}{2}$ विकल्प:
दिया गया है : x + y + z = 0
मान लीजिए x = 0
y = 1
z = –1
प्रश्न के अनुसार
$\Rightarrow \frac{x^{y} y^{2}+y^{2} z^{2}+z^{2} x^{2}}{x^{4}+y^{4}+z^{4}}=\frac{0+1+0}{0+1+1}=\frac{1}{2}$}}}}}}}}}}

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Todaysprint

If $x+y+z=0, \quad \frac{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}{x^{4}+y^{4}+z^{4}}=?$

If x+y+z=0,x2y2+y2z2+z2x2x4+y4+z4=?x+y+z=0, \quad \frac{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}{x^{4}+y^{4}+z^{4}}=?x+y+z=0,x4+y4+z4x2y2+y2z2+z2x2​=?

If $x+y+z=0, \quad \frac{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}{x^{4}+y^{4}+z^{4}}=?$