(x−2)3<−4x\frac{\left(x-2\right)}{3}<-4x3(x−2)<−4x
∫(u(u+1)3)du\int\left(\frac{u}{\left(u+1\right)^3}\right)du∫((u+1)3u)du
2x2 + 12x − 16 =02x^2\:+\:12x\:-\:16\:=02x2+12x−16=0
a−2=2a+ba-2=2a+ba−2=2a+b
16a54\sqrt[4]{16a^5}416a5
∫2(sin(x2))dx\int2\left(\sin\left(\frac{x}{2}\right)\right)dx∫2(sin(2x))dx
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