因式分解:x3+x2-y3-y2=

因式分解:x3+x2-y3-y2= 因式分解：x2-y2+x3-y3 x3-y3-x2+2xy-y2 因式分解 推导公式:(x+y)(x2-xy+y2)=x3+y3 因式分解X2(Y+Z)+Y2(Z+X)+Z2(X+Y)-(X3+Y3+Z3)-2XYZ 因式分解x3=8y3 (x-y)2(x2+xy+y2)2-(x3+y3)(-x3+y3),其中x=1,y=-1 解方程组：y2=x3-3x2+2x;x2=y3-3y2+2y2为平方 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, 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; 若x1/y1 =x2/y2 =x3/y3 =1/2,则（x1+x2-x3)/（y1+y2-y3）=? 在反比例y=-1/x的图像中,有三点（x1,y1),(x2,y2),(x3.y3),若x1>x2>0>x3,y1,y2,y3,的关系 证明：√[（x1-x2)^2+(y1-y2)^2]+√[(x2-x3)^2+(y2-y3)^2]>=√[(x1-x3)^2+(y1-y3)^2],x,y均为实数. x3-y3-x2+2xy-y2 因式分解还有：(2) x2+2xy-3y2+3x+y+2(3) 6x2-7x-4（实数范围内） x2+y2 分解因式已知x+y=10，x3+y3=100，求x2+y2的值 已知变量x1,x2,x3,由变量y1,y2,y3线性表示为：X1=2Y1+2Y2+Y3 X2=3Y1+Y2+5Y3 X3=3Y1+2Y2+3Y3试将变量Y1,Y2,Y3由变量X1,X2,X3表示 x3+6xy+y3 =x3+y3+6xy =(x+y)(x2-xy+y2)+6xy 为什么x3+y3+6xy可以化成(x+y)(x2-xy+y2)+6xy请为我解释一下恕我浅薄， 已知线性变换X1=2Y1+2Y2+Y3,X2=3Y1+Y2+5Y3,X3=3Y1+2Y2+3Y3,（三个式子用大括号括起）求从变量X1,X2,X3到变