−12∫0au12du-\frac{1}{2}\int_0^au^{\frac{1}{2}}du−21∫0au21du
u2−5u+2=0u^2-5u+2=0u2−5u+2=0
limx→6(e−6x4+e−3x)\lim_{x\to6}\left(\frac{e^{-6x}}{4+e^{-3x}}\right)x→6lim(4+e−3xe−6x)
x3−1x2−4\frac{x^3-1}{x^2-4}x2−4x3−1
x2−5x+8x+3\frac{x^2-5x+8}{x+3}x+3x2−5x+8
(0.4−5b)(0.4+5b)\left(0.4-5b\right)\left(0.4+5b\right)(0.4−5b)(0.4+5b)
ex+1y=xe^{x+1}y=xex+1y=x
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