$15-+-18$
$\frac{x^4}{121}+\frac{121}{16y^4}+\frac{1}{2}$
$3x^2=4x$
$1\:>\:2\:+\:3\:$
$a^2\cdot b+5a-3b;\:a=3;b=-1$
$2x^4-6x^2+4$
$2.6-5$
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