\(\lim\limits_{(x,y)\to(3,2)}(4x - 3y) =\)______ 。
举一反三
- \( \lim \limits_{(x,y) \to (2,1)} (4x - 3y) =\)______ 。
- 如下C程序的输出是什么？#include [stdio.h]void Func1 (int x, int y);void Func2 (int *x, int *y); int main() { int x = 3; int y = 4;Func1 (x, y); printf ("x = %d, y = %d\n", x, y);Func2(&x, &y); printf ("x = %d, y = %d\n", x, y);} void Func1 (int x, int y) { x = x + y; y = x - y; x = x - y; printf ("x = %d, y = %d\n", x, y);} void Func2 (int *x, int *y) { *x = *x + *y; *y = *x - *y; *x = *x - *y;;} A: x = 3, y = 4x = 3, y = 4x = 3, y = 4 B: x = 4, y = 3x = 4, y = 3x = 4, y = 3 C: x = 3, y = 4x = 3, y = 4x = 4, y = 3 D: x = 4, y = 3x = 3, y = 4x = 4, y = 3
- \(\lim \limits_{(x,y) \to (2,2)} { { x + y} \over {xy}} = \)______
- \( \mathop {\lim }\limits_{(x,y)______
- 应用Matlab软件计算行列式[img=110x88]17da5d7b00219d6.png[/img]为( ). A: x^2 - 6*x^2*y^2 + 8*x*y^3 - 3*y^4 B: x^3 - 6*x^2*y^2 + 8*x*y^3 - 3*y^4 C: x^4 - 6*x^2*y^2 + 8*x*y^3 - 3*y^4 D: x^5- 6*x^2*y^2 + 8*x*y^3 - 3*y^4