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Ответ:

[tex] \frac{2x {}^{2} }{1 + x {}^{4} } \leqslant 1 [/tex]

1+x⁴=0

x∉R

[tex] \frac{2x {}^{2} }{1 + x {}^{4} } \leqslant 1.x∉R[/tex]

[tex] \frac{2x {}^{2} }{1 + x {}^{4} } - 1 \leqslant 1 - 1[/tex]

[tex] \frac{2x {}^{2} }{1 + x {}^{4} } - 1 \leqslant 0[/tex]

[tex] \frac{2x {}^{2} }{1 + x {}^{4} } - \frac{1}{1} [/tex]

[tex] \frac{2x {}^{2} }{1 + x {}^{4} } - \frac{(1 + x {}^{4}) \times 1 }{(1 + x {}^{4} ) \times 1} [/tex]

[tex] \frac{2x {}^{2} }{ + x {}^{4} } - \frac{1 + x {}^{4} }{(1 + x {}^{4} ) \times 1} [/tex]

[tex] \frac{2x {}^{2} }{1 + x {}^{4} } - \frac{1 + x {}^{4} }{1 + x {}^{4} } [/tex]

[tex] \frac{2x {}^{2} - (1 + x {}^{4} )}{1 + x {}^{4} } [/tex]

[tex] \frac{2x { -}^{2} - (1 + x {}^{4} )}{1 + x {}^{4} } \leqslant 0[/tex]

[tex] \frac{2x {}^{2} - 1 - x {}^{4} }{1 + x {}^{4} } \leqslant 0[/tex]

-x⁴+2x²-1

-(x⁴-2x²+1)

[tex] \frac{-(x⁴-2x²+1)}{1 + x {}^{4} } \leqslant 0[/tex]

x^^^2*2-2x²+1

x^^^2*2-2*x²*1+1

x^^^2*2-2*x²*1+1²

[tex]a {}^{mn} = ( {a}^{n} ) {}^{m} [/tex]

(x²)²-2*x²*1+1²

a²-2ab+b²=(a-b)²

(x²-1)²

[tex] \frac{ - (x²-1)² }{1 + x {}^{4} } \leqslant 0[/tex]

[tex] - (x { } ^{2} - 1) {}^{2} \leqslant 0[/tex]

x∉R