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已知x,y,z成等差数列 求证:x2[y+z],y2[x+z],z2[x+y]成等差数列
人气:171 ℃ 时间:2020-04-18 15:32:40
解答
由x,y,z成等差数列,得x+z=2y
x^2(y+z)+z^2(x+y)=y(x^2+z^2)+xz(x+z)
=y((x+z)^2-2xz)+xz(x+z)
=y(4y^2-2xz)+2xyz
=4y^3-2xyz+2xyz
=4y^3
y^2(x+z)=y^2*2y=2y^3
所以,x^2(y+z)+z^2(x+y)=2y^2(x+z),即x^2[y+z],y^2[x+z],z^2[x+y]成等差数列
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