y′+5y=y23y'+5y=y^{\frac{2}{3}}y′+5y=y32
(34)4⋅(1.3)4⋅(83)2\left(\frac{3}{4}\right)^4\cdot\left(1.3\right)^4\cdot\left(\frac{8}{3}\right)^2(43)4⋅(1.3)4⋅(38)2
dudxx+y2=sin(y)\frac{du}{dx}\sqrt{x+y^2}=sin\left(y\right)dxdux+y2=sin(y)
limx→∞(5xx2+5)\lim_{x\to\infty}\left(\frac{5x}{\sqrt{x^2+5}}\right)x→∞lim(x2+55x)
ddxxx2+6(x+4)53\frac{d}{dx}\frac{x\sqrt{x^2+6}}{\left(x+4\right)^{\frac{5}{3}}}dxd(x+4)35xx2+6
(2x+3)(2x−4)\left(2x+3\right)\left(2x-4\right)(2x+3)(2x−4)
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