$\left(2-1\right)\cdot\left(2^2-2\right)^2$
$-3\left(-7+6\right)+3$
$3\left(1+1\right)^2-4$
$\frac{4}{sec\left(x\right)+tan\left(x\right)}$
$\ln\left(3x^2-4\right)+\ln\left(x^2+1\right)=\ln\left(2-x^2\right)$
$-1.02+-0.93+-0.30$
$\left(-4x+6y\right)^4$
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