问2道数学题,帮着解答,谢谢.

来源：百度知道 编辑：UC知道 时间：2024/05/15 10:51:14
1.(x^2-2x)^2-2x(2-x)+1 (因式分解)
2.已知m^2+n^2-6m+10n+34=0,求m,n的值
注x^y就是x的y次方.

1.(x^2-2x)^2-2x(2-x)+1
=(x^2-2x)^2-2(2x-x^2)+1
=(x^2-2x)^2+2(x^2-2x)+1
=(x^2-2x+1)^2

2.已知m^2+n^2-6m+10n+34=0,求m,n的值
(m^2-6m+9)+(n^2+10n+25)=0

(m-3)^2+(n+5)^2=0

m-3=0
n+5=0

m=3
n=-5

1.解：(x^2-2x)^2-2x(2-x)+1
=(x^2-2x)^2-2（2x-x^2)+1
=(x^2-2x)^2+2（x^2-2x)+1
=「(x^2-2x)+1」^2
=(x^2-2x+1)^2

2.解：m^2+n^2-6m+10n+34
=(m^2-6m+9)+(n^2+10n+25)+(34-9-25)
=(m-3)^2+(n+5)^2+0
∵m^2+n^2-6m+10n+34=0
∴(m-3)^2=0，(n+5)^2=0
∴m-3=0，n+5=0
∴m=3，n=-5

1、(x^2-2x)^2-2x(2-x)+1
＝【x（2-x）】^2-2*x*（2-x）+1
＝【x（2-x）-1】^2
＝（2x-x^2-1）^2
＝（x^2-2x+1）^2
＝（x-1）^4
2、m^2+n^2-6m+10n+34=0
m^2-6m+9+n^2+10n+25=0
(m-3)^2+(n+5)^2=0
m-3=0
n+5=0

m=3
n=-5