两题行列式题目| x 1 2 |1.| x x 2 || x 2 x || x^2 xy y^2 |2.| y^2 yz
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两题行列式题目
| x 1 2 |
1.| x x 2 |
| x 2 x |
| x^2 xy y^2 |
2.| y^2 yz z^2 |
| z^2 zx x^2 |
| x 1 2 |
1.| x x 2 |
| x 2 x |
| x^2 xy y^2 |
2.| y^2 yz z^2 |
| z^2 zx x^2 |
1、初等行变换R(2)-R(1); R(3)-R(1),易得x(x-1)(x-2)
2、=x^2y^2z^2*
I 1 y/x y^2/x^2 I
I 1 z/y z^2/y^2 I
I 1 x/z x^2/z^2 I
=(xyz)^2*
I 1 1 1I
Iy/x z/y x/zI
I(y/x)^2 (z/y)^2 (x/z)^2I
利用范德蒙公式
=(xyz)^2(z/y-y/x)(x/z-z/y)(x/z-y/x)
=(xz-y^2)(xy-z^2)(x^2-yz)
2、=x^2y^2z^2*
I 1 y/x y^2/x^2 I
I 1 z/y z^2/y^2 I
I 1 x/z x^2/z^2 I
=(xyz)^2*
I 1 1 1I
Iy/x z/y x/zI
I(y/x)^2 (z/y)^2 (x/z)^2I
利用范德蒙公式
=(xyz)^2(z/y-y/x)(x/z-z/y)(x/z-y/x)
=(xz-y^2)(xy-z^2)(x^2-yz)
(2X+Z-Y)/(X^2-XY+XZ-YZ)-(Y-Z)/(X^2-XY-XZ+YZ)
求(2X+Z-Y)/(X^2-XY+XZ-YZ)-(2X+Y+Z)/(X^2+XY+XZ+YZ)
12x^2yz-9x^2y^2 -x^2+xy-xz p(x-y)+q(y-x)
化简(2x-y-z/x^2-xy-xz+yz)+(2y-x-z/y^2-xy-yz+xz)+(2x-x-y/z^2-xz
xy/x+y=1,yz/y+z=2,xz/x+z=3求x,y
是关于因式分解的``1.计算x×x×x×x+5×x×x-6(x+y)(x+y+2xy)+(xy+1)(xy-1)2.已知
已知xy:yz:zx=3:2:1,求①x:y:z ②x/yz:y/zx
证明 (x+y+z)^2>3(xy+yz+zx)
求行列式第一行x^2+1 xy xz 第二行xy y^2+1 yz 第三行xz yz z^2+1
(xy-x^2)乘以(xy)/(x-y)
(x^2+xy/x-y)/(xy/x-y)计算
化简x^2-yz/[x^2-(y+z)x+yz]+y^2-zx/[y^2-(z+x)y+zx]+z^2-xy/[z^2-