• 2022-05-29
    y=In(x+根号下x^2+1)的奇偶性
  • 答:y=ln[x+√(x^2+1]定义域满足:x+√(x^2+1)>0√(x^2+1)>|x|>=-x恒成立所以:定义域为实数R,关于原点对称y(-x)=ln[-x+√(x^2+1)]=ln{1/[x+√(x^2+1)]}=-ln[x+√(x^2+1)]=-y(x)所以:y(x)是奇函数

    内容

    • 0

      解方程组:(根号3)*X—(根号2)*Y=1;(根号2)*X—(根号3)Y=0

    • 1

      x^2+y^2-4x-6y+13=0,求(2/3根号9x+y^2根号x/y^3)-x^2根号1/x-5x根号y/x)的值

    • 2

      方程${{x}^{2}}{{y}^{''}}-(x+2)(x{{y}^{'}}-y)={{x}^{4}}$的通解是( ) A: $y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$ B: $y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{4}})$ C: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{4}})$ D: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$

    • 3

      已知齐次方程$(x-1){{y}^{''}}-x{{y}^{'}}+y=0$的通解为$Y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}$,则方程$(x-1){{y}^{''}}-x{{y}^{'}}+y={{(x-1)}^{2}}$的通解是( ) A: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-({{x}^{2}}+1)$ B: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-({{x}^{3}}+1)$ C: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}$ D: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}+1$

    • 4

      积分递推公式1/(x^n根号(x^2+1))