Question

求几条初2数学题的答案

Answer 1

1.解不等式:
10-3(x+6)=1
-3(x+6)=-9
3(x+6)=9
(x+6)=3
x=-3
2.分解因式:
(3a+b)⒉-a⒉
=(3a+b+a)(3a+b-a)
=(4a+b)(2a+b)
3.计算:
x-2分之2x-6
÷
x⒉-4x+4分之x-3
=[(2x-6)/(x-2)]/[(x-3)/(x^2-4x+4)]
=[2(x-3)/(x-2)]*[(x+2)^2/(x-3)]
=2(x-2)
4.解方程:
x-3分之2=x分之3
2/(x-3)=x-3
2=(x-3)^2
±√2=x-3
x1=-(√2)+3(√代表根号)
x1=(√2)+3
5.有这样一道题“计算:
x⒉-1分之x⒉-2x+1除以x⒉+x分之x-1减x的值,其
中x=2007”.某同学把x=2007错抄成x=2008,但它的结果与正确答案相同,你说这是怎么回事?
x⒉-1分之x⒉-2x+1除以x⒉+x分之x-1减x
=[(x^2-2x+1)/(x^2-1)]/[(x-1)/(x^2+x)]-x
=[(x-1)^2/(x+1)(x-1)]*[x(x+1)/(x-1)]-x
=x-x
=0
6.利用分解因式说明:25⒎-5⒓能被24整除
证明=25^7-5^12
=(5^7)^2-5^12
=5^14-5^12
=5^12(5^2-1)
=5^12*24
所以25⒎-5⒓能被24整除
7.解不等式组:
5x-6≤2(x+3)
①
4分之x减1＜3分之x-3
②
化简①,得
x≤4
化简②,得
x>24
此题为空集(此题无解)