三道七年级上册数学题一、探究题设
1。a*b=a×b+a+b =ab+a+b,b*a=b×a+b+a =ab+a+b,a*b=b*a,
满足交换律 ,
(a+b)*c=(a+b)×c+(a+b)+c=ac+bc+a+b+c
a*c+b*c=(a×c+a+c)+(b×c+b+c)=ac+bc+a+b+2c,
(a+b)*c不等于a*c+b*c,不满足加法的分配律 。
2.1/(1×3)+1/(3×5)+1/(5×7)+……+1/(2n-1)(2n+1)
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+。。。+1/2(1/2n-1-1/2n+1)
=1/2[1-1/3+1/3-1/5+……+1/...全部
1。a*b=a×b+a+b =ab+a+b,b*a=b×a+b+a =ab+a+b,a*b=b*a,
满足交换律 ,
(a+b)*c=(a+b)×c+(a+b)+c=ac+bc+a+b+c
a*c+b*c=(a×c+a+c)+(b×c+b+c)=ac+bc+a+b+2c,
(a+b)*c不等于a*c+b*c,不满足加法的分配律 。
2.1/(1×3)+1/(3×5)+1/(5×7)+……+1/(2n-1)(2n+1)
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+。。。+1/2(1/2n-1-1/2n+1)
=1/2[1-1/3+1/3-1/5+……+1/(2n-1)-(2n+1)] (提取1/2)
=1/2(1-1/2n+1)
=n/(2n+1)
n/(2n+1)<1/2 第一项为1/3,所以一定大于1/3
3。
︱ x-2︱=1±a
x=2±(1±a)
x=3+a 或x=1+a 或x=3-a 或 x=1-a
3+a=1-a ,a=-1 ,或3+a=3-a,a=0
或1+a=3-a ,a=1 。
。收起