$\sqrt{3x+1}+1=x$
$\left(-9\right).\left(-2\right):\left(+3\right)$
$x^2+x^2+2x+1$
$\frac{1}{2}\cdot\left(x+4\right)\leqx$
$\left(+\frac{1}{3}mn\right)\cdot\left(+\frac{3}{2}m^2\right)$
$2\cdot256$
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