5+3x ≥4−x5+3x\:\ge4-x5+3x≥4−x
2cos2θ −1=02cos^2\theta\:-1=02cos2θ−1=0
(1−cos2b)(1+tan2b)\left(1-cos^2b\right)\left(1+tan^2b\right)(1−cos2b)(1+tan2b)
∫e3x1−3e3xdx\int\frac{e^{3x}}{1-3e^{3x}}dx∫1−3e3xe3xdx
8a2−16a−908a^2-16a-908a2−16a−90
(2−2)5\left(2-\sqrt{2}\right)^5(2−2)5
−(3x−2)2x+3-\frac{\left(3x-2\right)}{2x+3}−2x+3(3x−2)
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