举一反三
- (4)()#include()extern()serial_initial()();()main()(){()int()a,b();()unsigned()int()x,y();()serial_initial()();()a=b=0xaa55();x=y=0xaa55;()printf("\n()a=%4x()b=%4x()x=%4x()y=%4x",a,b,x,y)();()a=a<<1();b=b>>1;()x=x<<1();y=y>>1;()printf("\n()a=%4x()b=%4x()x=%4x()y=%4x",a,b,x,y)();()printf("\n")();()printf("\n")();()printf("That()is()all.\n")();()while(1)();()}()______________________________________________________________________________________________________________________________
- 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
- 已知函数f(x)是定义在(-∞,+∞)上的奇函数,当x∈[0,+∞)时,f(x)=x()2()-4x,则当x∈(-∞,0)时,f(x)=()(5.0分)A.()x()2()+4x()B.()x()2()-4x()C.()-x()2()+4x()D.()-x()2()-4x
- 已知\( y = {x^2} + 4x \)，则\( dy \)为（ ）. A: \( (2x + 4)dx \) B: \( 2xdx \) C: \( ({x^2} + 4)dx \) D: \( ({x^2} + 4x)dx \)
- 已知\( y = {x^2}{e^{ - x}} \)，则\( y'' \)为（ ）. A: \( 2{e^{ - x}} - 4x{e^{ - x}} - {x^2}{e^{ - x}} \) B: \( 2{e^{ - x}} - 4x{e^{ - x}} + {x^2}{e^{ - x}} \) C: 0 D: \( 2{e^{ - x}} - 4x{e^{ - x}} \)
内容
- 0
y﹦|4x-x²|去绝对值后是y=-x²+4,还是y=x²-4x?
- 1
将函数\(f(x)=\sin^4 x\)展开成Fourier级数为 ____ . A: \(f(x) = \frac{3}{8}-\frac{1}{2}\cos 2x +\frac{1}{8}cos 4x\) B: \(f(x) = \frac{1}{4}-\frac{1}{2}\cos x +\frac{3}{8}cos 4x\) C: \(f(x) = \frac{1}{4}-\frac{1}{2}\sin 2x -\frac{3}{8}cos 4x\) D: \(f(x) = \frac{3}{8}-\frac{1}{2}\sin x -\frac{1}{8}cos 4x\)
- 2
设$f(x)=x^2$,$g(x)=2^x$, 那么 $f \circ f \circ g=$ A: $2^{x^4}$ B: $2^{4x}$ C: $x^{2^{2x}}$ D: $x^{2^{x^2}}$
- 3
x＜-2-2＜x＜-1-1＜x＜33＜x＜4x＞4x+2-++++x+1--+++x-3---++x-4----+（x+2）（x+1）（x-3）（x-4）+-
- 4
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}]