从键盘输入一个实数x ,y=-x 2.5,
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![从键盘输入一个实数x ,y=-x 2.5,](/uploads/image/f/1779189-69-9.jpg?t=%E4%BB%8E%E9%94%AE%E7%9B%98%E8%BE%93%E5%85%A5%E4%B8%80%E4%B8%AA%E5%AE%9E%E6%95%B0x+%2Cy%3D-x+2.5%2C)
这个你用数形结合的方法很简单就可以得出答案,你去画一下他们的可行区域吧,我帮你做了一下答案是b=2
#include#includeusingnamespacestd;intmain(){floatx,y;cout
voidmain(){intx,inty;printf("请输入X的值:");scanf("%d",&x);printf("\n");if(x
publicstaticintfun(intx){if(x>0)return10;elseif(x==9)return1;elsereturn-10;}publicstaticvoidmain(Str
结合题目所给的不等式组,作出可行域【x-y+1≥0,x+y≥0,x≤0】,则x+2y在这个可行域内的最大值是2,最小值是0,则3^(x+2y)的最大值是3²=9,最小值是1.
#includevoidmain(){inta;printf("Inputthreenumbers:\n");scanf("%d",&a);printf("%d\n",a/10+a%10);print
由已知条件得:x2−1≥01−x2≥0x−1≠0,∴x=-1,y=3,∴y=(-1)3=-1.
#includevoidmain(){inta,b,c,t;printf("请输入3个数\n");scanf("%d%d%d",&a,&b,&c);if(b>c){if(a再问:哥哥输入xyz为什么程
input"x="toxdocasecasex
staticvoidMain(string[]args){inta=int.Parse(Console.ReadLine());intb=a%10;intc=(a/10)%10;intd=(a/100
x^=n; x的n次方不能这样写. ^在C里是异或操作#include <stdio.h>#include <math.h>int&nb
intfnX(intx){if(x>5)returnx+3;elseif(x>=0)return0;elsereturn2*x+30}main(){intdata;scanf("%d",&data);
importjava.util.*;publicclassCombination{publicstaticvoidmain(String[]args){VectortestData=newVector
PrivateSubCommand1_Click()i=Val(InputBox("inputanumber"))Ifi0Theny=x*23Elsey=-5*xx*2EndIfPrintyEndSu
#include#includeintmain(){floata,x,y;scanf("%f%f",&a,&x);y=pow(a,5)+sin(a*x)+log(a+x)+pow(2.718,a*x)
令t=2x+y,可得y=t-2x,代入x2+y24=1,得x2+14(t-2x)2=1化简整理,得2x2-tx+14t2-1=0∵方程2x2-tx+14t2-1=0有实数根∴△=t2-4×2×(14t
不知道你需要用什么语言,我就用C++给你写好了.#includeusingnamespacestd;intmain(){doublex;cin>>x;doubley;if(x=3&&x
当x<0时,25=(x+1)2,解得:x=-6,或x=4(舍去)当x≥0时,25=(x-1)2,解得:x=6,或x=-4(舍去)即输入的x值为±6故答案为:-6或6.
x=0y=0input"请输入一个实数"toxifx>0y=1ifx=0y=0elsey=-1endifendif