一道初一数学题已知：3x=6y=
1:已知：3x=6y=2z,且xy+yz+zx=99,求：
[2x^2-(x+y)(x-y)][(2-x)(x+2)+(-y-2)(2-y)]+z^2的值。
解：因为3x=6y=2z，所以，x=2y,z=3y,代入已知方程，得
2y×y+y×3y+3y×2y=99
11y^2=99
y^2=9
[2x^2-(x+y)(x-y)][(2-x)(x+2)+(-y-2)(2-y)]+z^2
=(2x^2-x^2+y^2)(4-x^2-4+y^2)+z^2
=(x^2+y^2)(y^2-x^2)+z^2
=y^4-x^4+z^2
=y^4-(2y)^4+(3y)^2
=y^4-16y^4+9y...全部
1:已知：3x=6y=2z,且xy+yz+zx=99,求：
[2x^2-(x+y)(x-y)][(2-x)(x+2)+(-y-2)(2-y)]+z^2的值。
解：因为3x=6y=2z，所以，x=2y,z=3y,代入已知方程，得
2y×y+y×3y+3y×2y=99
11y^2=99
y^2=9
[2x^2-(x+y)(x-y)][(2-x)(x+2)+(-y-2)(2-y)]+z^2
=(2x^2-x^2+y^2)(4-x^2-4+y^2)+z^2
=(x^2+y^2)(y^2-x^2)+z^2
=y^4-x^4+z^2
=y^4-(2y)^4+(3y)^2
=y^4-16y^4+9y^2
=9y^2-15(y^2)^2
=9×9-15×9^2
=-1134
2:3x=6y=2z 可得x:y:z = 2:1:3
设: x = 2k, y = k, z = 3k
xy+yz+zx=99 可得 11*k^2 = 99 可得 k^2 = 9
可得
x^2=4k^2=4*9=36, y^2=k^2=9, z^2=9k^2=9*9=81
可得
[2x^2-(x+y)(x-y)]*[(2-x)(x+2)+(-y-2)(2-y)]+z^2
=[2x^2-(x^2-y^2)](4-x^2+y^2-4)+z^2
=(2x^2-x^2+y^2)(y^2-x^2)+z^2
=(y^2+x^2)(y^2-x^2)+z^2
=(9+36)(9-36)+81
=81-1296+81=-1134
3:设3x=6y=2z=6m（最小公倍数6），则由xy+yz+zx=99推出m^2=9
而[2x^2-(x+y)(x-y)][(2-x)(x+2)+(-y-2)(2-y)]+z^2化简为y^4-x^4+z^2即
-15m^4+9m^2,代入m^2=9，可以得到结果-1134
。收起
