已知x,y,z互不相等,且xyz不等于0,x2+yz=z2,y2+zx=x2,求证:z2+xy=y2
人气:100 ℃ 时间:2019-08-22 18:09:57
解答
y2+zx=x2 => z=(x^2-y^2)/x代入x2+yz=z2=> x^4+xy(x^2-y^2)=(x^2-y^2)^2=> x^4+x^3y-xy^3=x^4+y^4-2x^2y^2=> x^3-xy^2-y^3+2x^2y=0z2^+xy-y^2=x^2+yz+xy-y^2=x^2+y(x^2-y^2)/x+xy-y^2=x^2-y^3/x+2xy-y^2=(...
推荐
- 已知2分之x=3分之y=4分之z,且xyz不等于0,求分式x2+y2+z2分之xy+yz+zx的值
- 已知x+y-z=0,2x-y-8z=0,且xyz不等于0,则x2+y2+z2/(xy+yz+zx)等于
- 若xyz属于r 求证 x2+y2+z2大于等于xy+yz+zx
- 已知xyz=1,x+y+z=2,x2+y2+z2=16?求[1÷(xy+2z)+1÷(yz+2x)+1÷(zx+2y)]的值
- 若xy+yz+zx=0,则3xy+x2(y+z)+y2(z+x)+z2(x+y)等于多少
- How to get along trouble dealing with some difficultt people 一篇120字左右的英语作文
- 请用act up ,act like .add up各造三个句子,
- 《长歌行》中,说明时光短暂易逝的句子是
猜你喜欢