用“可交换函数”的方法,这道题太简单了.
求证x^8+y^8+z^8>=x^3y^2z^2+x^2y^3z^2+x^2y^2z^3
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(1)3x+4y=15 2y-z=2 x+2z=9 (2) x+y+z=8 2x+3y-z=6 3x-y+z=8 (3)
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{x+2y+z=8,2x-y-z=-3,3x+y-2z=-1
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一.｛x+2y+3z=14 2x-y-z=-3 3x+y+z=8 二.｛3X-2Y-2Z=3 2X+2Y+3Z=12 X
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解方程组 x+2y+z=8 2x-y-z=-3 3x+y-2z=1
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{2x+y=3,3x-y+z=8,x+2y+z=13
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(1)2x+3y+z=17;(2)x+y+z=10;(3)3x+2y-z=8
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已知x，y，z∈R，且x+y+z=8，x2+y2+z2=24求证：[4/3]≤x≤3，[4/3]≤y≤3，[4/3≤z≤
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12x^2y^3z-8x^3y^2z+4x^2y
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2x-3y+4z=2 4x+y+7z=3 8x+2y+3z=-5
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x+y+z=-1 4x-2y+3z=5 y-z=8-2x