x1=4 y1=8,x2=8 y2=8 带入y=a（x-3）+k中a=?k=?

在LINGO中如何将整数输入min 100x1+100x2+40y1+40y2+40y3+40y4+40y5stx1+x2+y1>=4x1+x2+y1+y2>=3x1+x2+y1+y2+y3>=4x2+y1+y2+y3+y4>=6x1+y2+y3+y4+y5>=5x1+x2+y3+y4+y5>=6x1+x2+y4+y5>=8x1+x2+y5>=8y1+y2+y3+y4+y5 方程组y=ax+2 y²=8x的两组解为(x1,y1)(x2,y2)(x1≠x2)若x1+x2=4方程组y=ax+2 y²=8x的两组解为(x1,y1)(x2,y2)(x1≠x2)若x1+x2=4,求(x1-x2)²+（y1-y2)²的值. 这段lindo的代码有什么问题么~min = 100X1+100X2+40Y1+40Y2+40Y3+40Y4+40Y5STX1+X2+Y1>=4X1+X2+Y1+Y2>=3X1+X2+Y1+Y2+Y3>=4X2+Y1+Y2+Y3+Y4>=6X1+Y2+Y3+Y4+Y5>=5X1+X2+Y3+Y4+Y5>=6X1+X2+Y4+Y5>=8X1+X2+Y5>=8Y1+Y2+Y3+Y4+Y5=0@gin(x1)@gin(x2)@gin(y1)@gin(y2)@ PASCAL NOIP2008传纸条 关于重复数据处理（4维解法）重复处理部分为什么这样做?if (x1=x2)and(y1=y2) then f[x1,y1,x2,y2]:= f[x1,y1,x2,y2]+a[x1,y1]else f[x1,y1,x2,y2]:= f[x1,y1,x2,y2]+a[x1,y1]+a[x2,y2]; 已知等差数列 2,x1,x2,8和等比数列 2,y1,y2,8,则(x1+x2)/y1y2=? 过抛物线y^2=4x的焦点作直线,交抛物线于点A(x1,y1),B(x2,y2),若y1+y2=2√2,则|AB|的值为( )A.6 B.8 C.10 D.12我数学不好,算了半天我写出了这几步:F(1,0)∵y1^2=4x1 ,y2^2=4x2∴y1^2 - y2^2 = 4( x1 - x2)( y1 - y2 )( y1 + y2 数学公式推导max{|x1-x2|,|y1-y2|} =（|x1+y1-x2-y2|+|x1-y1-(x2-y2)|）/2这个公式是怎么推导出来的? 用配方法化二次型刘老师您好：f（x1,x2,x3）=x1^2+3x2^2+8x3^2+4x1x2+6x1x3+10x2x3配方后得f（x1,x2,x3）=（x1+2x2+3x3）^2-(x2+x3)^2 做可逆线性变换y1=x1+2x2+3x3 x1=y1-2y2-y3y2=x2+x3 即 x2=y2-y3y3=x3 x3=y3「即」前面的式 若数列a,x1,x2,b与数列a,y1,y2,y3,b均成等差数列(a≠b),则(x2-x1)/(y3-y2)注意（x2-x1)/(y3-y2）不是——（x2-x1)/(y2-y1)=因为等差数列所以b-x2=x2-x1=x1-a得到x2-x1=(b-a)/3而b-y3=y3-y2=y2-y1=y1-a得到y2-y1=(b-a)/4所以（x2 已知A(x1,y1),B(x2,y2)是抛物线y2=4x上的任意2点,若y1y2=-8,则直线AB过定点------ 杭电2001.为什么是wrong answer?#include#includemain(){float x1,x2,y1,y2;while(scanf(%f %f %f %f,&x1,&x2,&y1,&y2)!=EOF){printf(%.2f,sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)));putchar(' ');}} 已知a,x1,x2,b成等差数列,a,y1,y2,b成等比数列,则x1+x2+y1×y2=? matlab如何将一组数代入原方程的字母里?syms x1 x2 x3 x4 y1 y2 y3 y4 X1 X2 X3 X4 Y1 Y2 Y3 Y4A=[x1,y1,1,0,0,0,-x1*X1,-y1*X1;0,0,0,x1,y1,1,-x1*Y1,-y1*Y1;x2,y2,1,0,0,0,...-x2*X2,-y2*X2;0,0,0,x2,y2,1,-x2*Y2,-y2*Y2;x3,y3,1,0,0,0,-x3*X3,-y3*X3; 两点式方程(y-y1)/(y2-y1)=(x-x1)/(x2-x1)怎么证出 过抛物线x2=4y的焦点F作直线交抛物线于P1(X1,Y1),P2(X2,Y2)俩点,若Y1+Y2=6则∣P1P2∣的值为 A.5 B.6 C.8 过抛物线x2=4y的焦点F作直线交抛物线于P1(X1,Y1),P2(X2,Y2)俩点,若Y1+Y2=6则∣P1P2∣的值为A.5 B.6 C.8 D.10 已知点A(X1,Y1)B(X2,Y2)均在反比例函数Y=-2/X的图像上且X1乘X2=8,求Y1乘Y2的值 已知a、b、c为已知数,x1、x2、y1、y2未未知数,有如下关系式：------------------------------------------------------------------------------x1 + x2 = C;x1 * a - x2 = y1;x2 * b - x1 = y2;求“y1 + y2达到最大值时,x1与x2的值 高中数学关于斜率的问题,应该是知识点设A（X1,y1）B（X2,y2）代入双曲线则X1^2-4y1^2=4X2^2-4y2^2=4两式相减（X1+X2）（X1-X2）-4（y1+y2）(y1-y2）=0则k=（y1-y2）/（X1-X2）=（X1+X2）/4（y1+y2）