【初2】几道数学题目````高手速度做答```````1'分解固式(1-a)mn+a-1=2'若18x^2+19x+m=(9x+5)(2x+n),则m=?n=?3'若关于x的三次二项式2x^2+mx+n因式分解的结果为(2x-1)(x+1/4),求m,n的植.4'两个小孩的年龄分别是x,y,已知x^2+xy=99,求这两个孩子的年龄.(过程)尽量让我懂哈````!过程`````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````

1
提取公因式法
(1-a)mn+a-1
=(1-a)mn-(1-a)
=(1-a)(mn-1)
2
18x^2+19x+m
=(9x+5)(2x+n)
=18x^2+(10+9n)x+5n
等号两边对应的各项系数及常数项应相等
所以
19=10+9n
m=5n
解得n=1,m=5
3
2x^2+mx+n 这是个二次三项式，不是三次二项式，解法跟第二题一模一样，我就不写了，留给你作练习
4
很容易观察出，一个小孩9岁，另一个11岁。没有过程

1.(mn-1)(1-a)
2.m=5 n=1
3.将(2x-1)(x+1/4)乘开，得2x^2-1/2x-1/4,所以m=-1/2 n=-1/4
4.岁数是整数,且不为负，所以式子化为x(x+y)=11*9，易解地，x=9,y=2

1
提取公因式法
(1-a)mn+a-1
=(1-a)mn-(1-a)
=(1-a)(mn-1)
2
18x^2+19x+m
=(9x+5)(2x+n)
=18x^2+(10+9n)x+5n
等号两边对应的各项系数及常数项应相等
所以
19=10+9n
m=5n
解得n=1,m=5
3
2x^2+mx+n 这是个二次三项式，不是三次二项式，解法跟第二题一模一样，我就不写了，留给你作练习
4
很容易观察出，一个小孩9岁，另一个11岁。没有过程!
哈哈

1.(1-a)(mn-1)
2.18x^2+19x+m=(9x+5)(2x+n)=18x^2+(9n+10)x+5n
9n+10=19
5n=m
m=5
n=1
3.同题二分解得出m=-3/2,n=-1/4
4.x^2+xy=x(x+y)=99=9*(9+2)
x=9,y=2

1.提取公因式(1-a)mn+a-1 =(1-a)mn-(1-a) =(1-a)(mn-1) 2 .18x^2+19x+m =(9x+5)(2x+n) =18x^2+(10+9n)x+5n 等号两边对应的各项系数及常数项相等 所以 :19=10+9n m=5n 解得n=1,m=5 3 .(2x-1)(x+1/4)=2x^2+x*1/2-x-(1/...