求(1-x)^2019的积分
A: -1/2020(1-x)^2020+C
B: 2020(1-x)^2020+C
C: -1/2019(1-x)^2020+C
D: 1/2019(1-x)^2020+C
A: -1/2020(1-x)^2020+C
B: 2020(1-x)^2020+C
C: -1/2019(1-x)^2020+C
D: 1/2019(1-x)^2020+C
A
举一反三
- 函数y=1-x2(x<0)的反函数为( ) A: y=1-x(x<1) B: y=-1-x(x≤1) C: y=-1-x(x<1) D: y=1-x(x≤1)
- 求极限lim(x→1)sin(1-x)/(1-x^2)
- 已知A={1,x,-1},B={-1,1-x},
- F(1/X)=X/1-X*X求f(x)=
- 17e0b849b7d64bd.jpg,计算[img=19x34]17e0ab14a855463.jpg[/img]实验命令为(). A: syms x;f=diff(asinsqrt(x))f=1/2/x^(1/2)/(1-x)^(1/2) B: f=diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2) C: syms x;diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2)
内容
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17da42840675a6d.jpg,计算[img=19x34]17da4275482315f.jpg[/img]实验命令为(). A: syms x;f=diff(asinsqrt(x))f=1/2/x^(1/2)/(1-x)^(1/2) B: f=diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2) C: syms x;diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2)
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若f′(cos2x)=sinx,则f(x)等于:() A: (1/3)(1-x)+c B: (2/3)(1-x)+c C: -(1/3)(1-x)+c D: (1-x)+c
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y=ln(1-x²)的定义域为: -1≤x≤1|-1<x<1|0≤x≤1|0<x<1
- 3
已知x不等于1,计算(1+x)(1-x)=1-x^2,(1-x)(1+x+x^2)=1-x^3,(1-x)(1+x+x^2+x^3)=1-x^4
- 4
下面级数求和错误的是 A: $\sum_{n=0}^\infty q^n = \frac{1}{1-q} (0\lt q\lt1) $ B: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{x}{1-x} (|x|\lt 1) $ C: $\sum_{n=1}^\infty \frac{1}{{n!}} = e $ D: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{1}{1-x} (x>1) $