∣1+(6−9)∣−(8−12)\left|1+\left(6-9\right)\right|-\left(8-12\right)∣1+(6−9)∣−(8−12)
dydx=(z−1)2\frac{dy}{dx}=\left(z-1\right)^2dxdy=(z−1)2
(sec(y)−tan(y))(sec(y)+tan(y))sec(y)=cos y\frac{\left(sec\left(y\right)-tan\left(y\right)\right)\left(sec\left(y\right)+tan\left(y\right)\right)}{sec\left(y\right)}=cos\:ysec(y)(sec(y)−tan(y))(sec(y)+tan(y))=cosy
x3+8x2−5x+10x−2\frac{x^3+8x^2-5x+10}{x-2}x−2x3+8x2−5x+10
∫(2x2−4x)6(x−1)dx\int\left(2x^{2}-4x\right)^{6}\left(x-1\right)dx∫(2x2−4x)6(x−1)dx
limx→0(1+cos(x)x2sinx)\lim_{x\to0}\left(\frac{1+\cos\left(x\right)}{x^2sinx}\right)x→0lim(x2sinx1+cos(x))
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