若x+y+z=0且x,y,z互不相等.求x^2/(2x^2+yz)+y^2/(2y^+xz)+z^2/(2z^2+xy).在线等.
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![若x+y+z=0且x,y,z互不相等.求x^2/(2x^2+yz)+y^2/(2y^+xz)+z^2/(2z^2+xy).在线等.](/uploads/image/z/1577254-22-4.jpg?t=%E8%8B%A5x%2By%2Bz%3D0%E4%B8%94x%2Cy%2Cz%E4%BA%92%E4%B8%8D%E7%9B%B8%E7%AD%89.%E6%B1%82x%5E2%2F%282x%5E2%2Byz%29%2By%5E2%2F%282y%5E%2Bxz%29%2Bz%5E2%2F%282z%5E2%2Bxy%29.%E5%9C%A8%E7%BA%BF%E7%AD%89.)
若x+y+z=0且x,y,z互不相等.求x^2/(2x^2+yz)+y^2/(2y^+xz)+z^2/(2z^2+xy).在线等.
若x+y+z=0且x,y,z互不相等.求x^2/(2x^2+yz)+y^2/(2y^+xz)+z^2/(2z^2+xy).在线等.
若x+y+z=0且x,y,z互不相等.求x^2/(2x^2+yz)+y^2/(2y^+xz)+z^2/(2z^2+xy).在线等.
假设x=0 y=1 z=-1
则:
x^2/(2x^2+yz)+y^2/(2y^+xz)+z^2/(2z^2+xy)
=0/(0-1)+1/(2-0)+(-1)²/(2+0)
=0+1/2+1/2
=1
1)数值带入法,a=1,b=-1,c=0;a=1,b=2,c=-3.带入两组数据后结果为1
2)z=-x-y,代入得x^2/(2x^2-xy-y^)+y^2/(2y^2-xy-x^)+(x+y)^2/(2X^2+5xy+2y^2),分母进行多项式因式分解前两项进行通分进行加法计算有
(x-y)(x^2+3xy+y^2)/[(x-y)(2x+y)(x+2y)]+(x+y)^2/[(2x+y)(x+2y)]
最后化简为1