已知x,y,z满足x/(y+z)+y/(z+x)+z/(x+y)=1,求代数式x2/(y+z)+y2/(x+z)+z2/
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/09/20 08:00:05
已知x,y,z满足x/(y+z)+y/(z+x)+z/(x+y)=1,求代数式x2/(y+z)+y2/(x+z)+z2/(x+y)的值、
2是平方
2是平方
x/(y+z)+y/(z+x)+z/(x+y)=1
所以x/(y+z)=1-[y/(z+x)+z/(x+y)]
y/(z+x)=1-[x/(y+z)+z/(x+y)]
z/(x+y)=1-[x/(y+z)+y/(z+x)]
x²/(y+z)+y²/(z+x)+z²/(x+y)
=x*[x/(y+z)]+y*[y/(z+x)]+z*[z/(x+y)]
=x*{1-[y/(z+x)+z/(x+y)]}+y*{1-[x/(y+z)+z/(x+y)]}+z*{1-[x/ (y+z)+y/(z+x)]}
=x-x*[y/(z+x)+z/(x+y)]+y-y*[x/(y+z)+z/(x+y)]+z-z*[x/(y+z)+y/(z+x)]
=x+y+z-[xy/(z+x)+xz/(x+y)+yx/(y+z)+yz/(x+y)+zx/(y+z)+zy/(z+x)]
=x+y+z-[(xy+zy)/(z+x)+(yx+zx)/(y+z)+(xz+yz)/(x+y)]
=x+y+x-[y(x+z)/(z+x)+x(y+z)/(y+z)+z(x+y)/(x+y)]
=x+y+z-(y+x+z)
=0
所以x/(y+z)=1-[y/(z+x)+z/(x+y)]
y/(z+x)=1-[x/(y+z)+z/(x+y)]
z/(x+y)=1-[x/(y+z)+y/(z+x)]
x²/(y+z)+y²/(z+x)+z²/(x+y)
=x*[x/(y+z)]+y*[y/(z+x)]+z*[z/(x+y)]
=x*{1-[y/(z+x)+z/(x+y)]}+y*{1-[x/(y+z)+z/(x+y)]}+z*{1-[x/ (y+z)+y/(z+x)]}
=x-x*[y/(z+x)+z/(x+y)]+y-y*[x/(y+z)+z/(x+y)]+z-z*[x/(y+z)+y/(z+x)]
=x+y+z-[xy/(z+x)+xz/(x+y)+yx/(y+z)+yz/(x+y)+zx/(y+z)+zy/(z+x)]
=x+y+z-[(xy+zy)/(z+x)+(yx+zx)/(y+z)+(xz+yz)/(x+y)]
=x+y+x-[y(x+z)/(z+x)+x(y+z)/(y+z)+z(x+y)/(x+y)]
=x+y+z-(y+x+z)
=0
已知x,y,z满足x/(y+z)+y/(z+x)+z/(x+y)=1,求代数式x2/(y+z)+y2/(x+z)+z2/
已知实数x,y,z满足x/(y+z)+y/(z+x)+z/(x+y)=1,求x2/(y+z)+y2/(z+x)+z2/(
已知:(x+y-z)/z=(x-y+z)/y+(y+z-x)/x,且xyz≠0,求代数式[(x+y)(y+z)(x+z)
已知xyz=1,x+y+z=2,x2+y2+z2=16,求1/x+y+1/y+z+1/x+z
已知x+y+z=1,x2+y2+z2=2,x3+y3+z3=3,求xy(x+y)+yz(y+z)+zx(z+x)的值
已知x+y+z=1,xy+yz+zx=2,xyz2,求x2(y+z)+y2(z+x)+z2(x+y)的值
已知x、y、z满足x2-4x+y2+6y+z+1
已知实数x,y,z满足以下条件,求x的取值范围.x+y+z=a,x2+y2+z2=1/2 a2
1.已知x2+y2+z2-2x+4y-6z+14=0,求x+y+z的值.
已知x2+4y2+z2-2x+4y-6z+11=0 求x+y+z的值
已知xyz=1,x+y+z=2,x2+y2+z2=16,求代数式1xy+2z+1yz+2x+1zx+2y
(x+y-z)(x-y+z)=