(1)已知x+2z=3y,试判断x^2-9y^2+4z^2+4xz的值是不是定值,求出它的值,否则请说明理由(2)已知x^2-2x=2,将下式先化简,再求值:(x-1)^2+(x-3)(x+3)+(x-3)(x-1)
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(1)已知x+2z=3y,试判断x^2-9y^2+4z^2+4xz的值是不是定值,求出它的值,否则请说明理由
(2)已知x^2-2x=2,将下式先化简,再求值:(x-1)^2+(x-3)(x+3)+(x-3)(x-1)
(1)它的值是个定值,定值为0
等式两边平方可得,
(x+2z)^2=(3y)^2
x^2+4xz+4z^2=9y^2
x^2+4xz+4z^2-9y^2=0
原式=x^2-2x+1+x^2-9+x^2-4x+3
=3x^2-6x-5
=3(x^2-2x)-5
=3×2-5
=1
ps:liu30003000和张俊逸的第二题答案有两个错误
(1)倒数第三行到倒数第二行,第一个数字3移下来变成了2
=3(x^2-2x+1)-8
=2(x-1)^2-8
=10
(2)倒数第二行(x-1)^2=3带入,而这两个人带入了3以后又平方了一次,导致带入的是x-1=3,错误.
1。x^2-9y^2+4z^2+4xz=(x+2z)^2-9y^2=0是定值。
2。x^2-2x=2,(x-1)^2=3
(x-1)^2+(x-3)(x+3)+(x-3)(x-1)
=x^2-2x+1+x^2-9+x^2-4x+3
=3x^2-6x-5
=3(x^2-2x+1)-8
=2(x-1)^2-8
=10
1。x^2-9y^2+4z^2+4xz=(x+2z)^2-9y^2=0是定值。
2。x^2-2x=2,(x-1)^2=3
(x-1)^2+(x-3)(x+3)+(x-3)(x-1)
=x^2-2x+1+x^2-9+x^2-4x+3
=3x^2-6x-5
=3(x^2-2x+1)-8
=2(x-1)^2-8
=10
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