初一数学题

显示全部楼层 · 2009-4-5 21:31:26

因式分解：（x-1)(x-y-z-1)+yz
已知a,b,c满足:
a的平方+2b=7;b的平方-2c=-1;c的平方-6a=-17
求a+b+c
要详细过程，谢谢

千问 · 2009-4-5 21:31:26

（x - 1)(x - y - z - 1) + yz= （x - 1)[(x - 1) - (y + z)] + yz= （x - 1)^2 - (y + z)(x - 1) + yz = (x - 1 - y)(x - 1 - z)(此步实际就是十字相乘法,把(x - 1)看成一个整体)a^2 + 2b = 7b^2 - 2c = -1c^2 - 6a = -17 三式相加得:a^2 - 6a + b^2 + 2b + c^2 - 2c = -11(a - 3)^2 + (b +1)^2 + (c - 1)^2 = 0a = 3b = -1c = 1则 a ...

千问 · 2009-4-5 21:31:26

（x-1)(x-y-z-1)+yz =(x-1)[(x-1)-(y+z)]+yz=(x-1)^2-(y+z)(x-1)+yz=(x-1-y)(x-1-z)a的平方+2b=7;b的平方-2c=-1;c的平方-6a=-17 三式相加a^2-6a+b^2+2b+c^2-2c+11=0∴(a-3)^2+(b+1)^2+(c-1)^2=0...