x+y+z=0 求证:(x^2-y^2)+(xz-yz)=0
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/19 05:25:44
x)ЮԮ5Px{fiTVijkTTVVi$PΆ31@S4l4amByvP{m}g.&m0Ybv@>t2>ӁfVXOvz,@[ݪ{z^u)ąt0́j~o:=H ޢY
x+y+z=0 求证:(x^2-y^2)+(xz-yz)=0
x+y+z=0 求证:(x^2-y^2)+(xz-yz)=0
x+y+z=0 求证:(x^2-y^2)+(xz-yz)=0
(x^2-y^2)+(xz-yz)
=(x-y)(x+y)+z(x-y)
=(x-y)(x+y+z)
=0
原式=(x+y)(x-y)+(x-y)z=(x-y)(x+y+z)=0
证明:
(x^2-y^2)+(xz-yz)
=(x+y)(x-y)+(x-y)*z
因为x+y+z=0
所以x+y=-z,代入原式
=-z*(x-y)+(x-y)*z
=0
得证
x>=y>=z>0,求证:x^2*y/z+y^2*z/x+z^2*x/y>=x^2+y^2+z^2
已知x-y/x+y=y+z/2(y-z)=z+x/3(z-x),求证8x+9y+5z=0THX..
已知x,y,z 大于0,x+y+z=2,求证 xz/y(y+z)+zy/x(x+y)+yx/z(z+x)大于等于2/3
已知x,y,z>0,xyz(x+y+z)=1,求证(x+y)(x+z)>=2
(z-x)^2-4(x-y)(y-z)=0求证x+z-2y=0_
若(z-x)的二次方-4(x-y)(y-z)=0,求证x-2y+z=0
已知(x-z)^2-4(x-y)(y-z)=0,求证:2y=x+z
若x-2y+z=0 求证(x-z)^2-4(x-y)(y-x)=0
实数x,y,z满足(x-z)的平方减4(x-y)(y-z)=0求证z+x-2y=0
已知三个实数x,y,z满足条件(z-x)^2-4(x-y)(y-z)=0,求证:x,y,z成等差数列
x,y,z∈(0,1),且x+y+z=2,求证1
X/(Y-Z)+Y/(Z-X)+Z/(X-Y)=0求证X/(Y-Z)^2+Y/(Z-X)^2+Z/(X-Y)^2=0通过上面求证下面^2是……的2次方X/(Y-Z)^2+Y/(Z-X)^2+Z/(X-Y)^2=0
已知:x^2/(z+y)+y^2/(x+z)+z^2/(x+y)=0,求x/(z+y)+y/(x+z)+z/(x+y)的值.
x^2/(z+y)+y^2/(x+z)+z^2/(x+y)=0,求x/(z+y)+y/(x+z)+z/(x+y)的值
已知x+y+z=0求证x*x*x+y*y*y+z*z*z=3xyz
若xyz=1,求证 x^2/(y+z)+y^2/(z+x)+z^2/(x+y)≥3/2
x,y,z是实数,且x^2+y^2+z^2=2,求证:x+y+z
若x>y>z>0,求证:x-z+27/[(x+z)y-xz-y^2]≥9