x1+x2+x3=y1+y2+y3=x1y1+x2y2+x3y3=0,求x1^2/(x1^2+x2^2+x3^2)+y1^2/(y1^2+y2^2+y3^2)的取值范围
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已知A(x1,y1),B(x2,y2),C(x3,y3),求二次函数的a b c找到了a=(y3)/(((x2)-(x3))*((x1)-(x3)))+(y2)/(((x2)-(x3))*((x2)-(x1)))+(y1)/(((x1)-(x2))*((x1)-(x3)));b=((y2)-(y1)-a*((x2)*(x2)-(x1) *(x1)))/((x2)-(x1));c=y1-a*(x1)*(x1)-b*x1;
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;
MATLAB 已知3点求夹角>> x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;theta=acosd(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,y1-y2])*norm([x1-x2,y3-y2])))theta =47.8696>> x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;theta=acos(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,
点(x1,y1),(x2,y2),(x3,y3)其中x1
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x1+x2+x3=y1+y2+y3=x1y1+x2y2+x3y3=0,求x1^2/(x1^2+x2^2+x3^2)+y1^2/(y1^2+y2^2+y3^2)的取值范围
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