设圆C与两圆(x+√5)^2+y^2=4,(x-√5)+y^2=4中的一个内切,另一个外切.圆心
举一反三
- 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: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$
- 下面程序的正确输出结果是( )。 #includemain(){ int x=10,y=5; switch(x) { case 1:x+y; default:x-=y; case 2:y--; case 3:x--; } printf("x=%d,y=%d",x,y);} A: x=4,y=4 B: x=10,y=5 C: x=5,y=5 D: x=5,y=4
- 设x=2,y=3,z=4,k=5，则表达式“x 2
- 求解方程组[img=218x63]1803072f0e0e849.png[/img]接近 (2,2) 的解 A: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] B: NSolve[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] C: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,y},{2,2}] D: FindRoots[{x^2+y^2=5Sqrt[x^2+y^2]-4x,y=x^2},{x,2},{y,2}]
- 求解方程组[img=218x63]1803072e5daced1.png[/img]接近 (2,2) 的解 A: NSolve[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] B: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] C: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,y},{2,2}] D: FindRoots[{x^2+y^2=5Sqrt[x^2+y^2]-4x,y=x^2},{x,2},{y,2}]