• 2022-06-15 问题

    设"f:R→R, f(x)"=x^2-x+2; "g:R→R," "g(x)"=x-3"," 则"f°g(x)="( )

    设"f:R→R, f(x)"=x^2-x+2; "g:R→R," "g(x)"=x-3"," 则"f°g(x)="( )

  • 2022-06-03 问题

    由曲线y=x^3-2x^2-x+2与x轴所围成平面图形的面积为?

    由曲线y=x^3-2x^2-x+2与x轴所围成平面图形的面积为?

  • 2022-06-09 问题

    设\(z = \int_ { { x^2}}^y { { e^t}\sin t} dt\),则\({z_{xx}=}\) A: \(2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} + 2{x^2}\cos {x^2}} \right]\) B: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} - 2{x^2}\cos {x^2}} \right]\) C: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} + 2{x^2}\cos {x^2}} \right]\) D: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\cos {x^2} + 2{x^2}\sin {x^2}} \right]\)

    设\(z = \int_ { { x^2}}^y { { e^t}\sin t} dt\),则\({z_{xx}=}\) A: \(2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} + 2{x^2}\cos {x^2}} \right]\) B: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} - 2{x^2}\cos {x^2}} \right]\) C: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} + 2{x^2}\cos {x^2}} \right]\) D: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\cos {x^2} + 2{x^2}\sin {x^2}} \right]\)

  • 2022-06-07 问题

    求不定积分[img=132x48]17da6537fc8dad6.png[/img]; ( ) A: -(4*(cos(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) B: (4*(sin(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) C: (4*(cos(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) D: (4*(cos(x/2)/2 + 2*cos(x/2)))/(17*exp(2*x))

    求不定积分[img=132x48]17da6537fc8dad6.png[/img]; ( ) A: -(4*(cos(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) B: (4*(sin(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) C: (4*(cos(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) D: (4*(cos(x/2)/2 + 2*cos(x/2)))/(17*exp(2*x))

  • 2022-06-19 问题

    与数学关系式[img=114x22]1803bce8722f322.png[/img]等价的C语言关系表达式是? A: x < -2 && x > 2 B: x < -2 || x > 2 C: -2 < x < 2 D: !(-2 <= x <=2) E: !(-2 <=x && x <= 2) F: x < -2, x > 2

    与数学关系式[img=114x22]1803bce8722f322.png[/img]等价的C语言关系表达式是? A: x < -2 && x > 2 B: x < -2 || x > 2 C: -2 < x < 2 D: !(-2 <= x <=2) E: !(-2 <=x && x <= 2) F: x < -2, x > 2

  • 2022-06-11 问题

    求函数[img=148x49]17da6537a5eee98.png[/img]的导数; ( ) A: 1/(x^2*(2/x^2 + 1)) B: -1/(x^2*(2/x^2 + 1)) C: (x^2*(2/x^2 + 1)) D: -1/(x^2*(2/x^2 + 1))+2/x^2 + 1

    求函数[img=148x49]17da6537a5eee98.png[/img]的导数; ( ) A: 1/(x^2*(2/x^2 + 1)) B: -1/(x^2*(2/x^2 + 1)) C: (x^2*(2/x^2 + 1)) D: -1/(x^2*(2/x^2 + 1))+2/x^2 + 1

  • 2022-06-03 问题

    求函数[img=102x46]17da6537bc771a0.png[/img]的导数; ( ) A: -x/(a^2 - x^2)^(3/2) B: x/(a^2 - x^2)^(3/2) C: (a^2 - x^2)^(3/2)/x D: (a^2 - x^2)^(1/2)

    求函数[img=102x46]17da6537bc771a0.png[/img]的导数; ( ) A: -x/(a^2 - x^2)^(3/2) B: x/(a^2 - x^2)^(3/2) C: (a^2 - x^2)^(3/2)/x D: (a^2 - x^2)^(1/2)

  • 2022-06-01 问题

    求微分方程[img=101x35]17da5f15503f795.png[/img] 的通解,实验命令为(). A: dsolve(Dy+2*x*y=x*exp(-x^2))ans=C1*exp(-x^2) + (x^2*exp(-x^2))/2 B: dsolve('Dy+2*x*y=x*exp(-x^2)','x')ans=C1*exp(-x^2) + (x^2*exp(-x^2))/2 C: dsolve('Dy+2*x*y=x*exp(-x^2)')ans=C1*exp(-x^2) + (x^2*exp(-x^2))/2

    求微分方程[img=101x35]17da5f15503f795.png[/img] 的通解,实验命令为(). A: dsolve(Dy+2*x*y=x*exp(-x^2))ans=C1*exp(-x^2) + (x^2*exp(-x^2))/2 B: dsolve('Dy+2*x*y=x*exp(-x^2)','x')ans=C1*exp(-x^2) + (x^2*exp(-x^2))/2 C: dsolve('Dy+2*x*y=x*exp(-x^2)')ans=C1*exp(-x^2) + (x^2*exp(-x^2))/2

  • 2022-06-19 问题

    与数学关系式[img=114x22]17de85f03ec5890.png[/img]等价的C语言关系表达式是? A: x <; -2 &amp;&amp; x >; 2 B: x <; -2 || x >; 2 C: -2 <; x <; 2 D: !(-2 <;= x <;=2) E: !(-2 <;=x &amp;&amp; x <;= 2) F: x <; -2, x >; 2

    与数学关系式[img=114x22]17de85f03ec5890.png[/img]等价的C语言关系表达式是? A: x <; -2 &amp;&amp; x >; 2 B: x <; -2 || x >; 2 C: -2 <; x <; 2 D: !(-2 <;= x <;=2) E: !(-2 <;=x &amp;&amp; x <;= 2) F: x <; -2, x >; 2

  • 2022-05-31 问题

    数学式 A: (e^(2*x)*Log(x)+x^2)/Sqr(Abs(Sinx^2-Cos2x)) B: (Exp(2*x)*Log(x)+x^2)/Sqr(Abs(Sin(x^2)-Cos(x)^2)) C: (Exp(2*x)*Ln(x)+x^2)/Sqr(Abs(Sin(x^2)-Cos(x)^2)) D: (e^(2*x)*Log(x)+x^2)/Sqr(Abs(Sin(x)^2-Cos(x)^2))

    数学式 A: (e^(2*x)*Log(x)+x^2)/Sqr(Abs(Sinx^2-Cos2x)) B: (Exp(2*x)*Log(x)+x^2)/Sqr(Abs(Sin(x^2)-Cos(x)^2)) C: (Exp(2*x)*Ln(x)+x^2)/Sqr(Abs(Sin(x^2)-Cos(x)^2)) D: (e^(2*x)*Log(x)+x^2)/Sqr(Abs(Sin(x)^2-Cos(x)^2))

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