tfsf.net
当前位置:首页>>关于观察下列一组式子(√2+1)(√2-1)=1, (√3+√2)(√3-√2)=...的资料>>

观察下列一组式子(√2+1)(√2-1)=1, (√3+√2)(√3-√2)=...

观察下列一组式子:(√2+1)(√2-1)=1, (√3+√2)(√3-√第1小题用分母有理化解决 比如1/根号2+1=(根号2-1)/1,可以发现后面很多可以消掉 第二小题可先比较根号11+根号10与根号12+

1)=(√2)²-1=1;(√3+√2)(√3-√2)=(√3)&#(√n+√(n-1))(√n-√(n-1))=1这是个平方差公式,利用平方差公式可直接得证.

下列式子:(1+√2)×(√2-1)=1 (√2+√3)×(√3-√2)=1[1/(1+√2)+1/(√2+√3)+1/(√3+√4)+1/(√4+√5)+……+1/(√2012+√2013)]×(1+√2013)=(√2-1+√3-

观察下列一组等式,(根号2+1)(根号2-1)=1;(根号3+根号2[1/(√2+1)+1/(√3+√2)+1/(√4+√3)+````+1/(√2012+√2011)](√2012+1

2-1)=1;(根号3+根号2)(根号3-根号2)=1;(根号4+根号3/(√2+1)(√2-1)+(√3-√2)/(√3+√2)(√3-√2)+(√2012-√2011)/(√2012+√2011)(√2012-√2011)](√

=√2-1,②1/√3+√2=√3-√2/(√3+√2)(√3-√2)=(1)规律为:1/[√n+√(n-1)]=√n-√(n-1)(2)1/(√2+1)+1/(√3+√2+1/(

/√2+1=√2-1/(√2+1)(√2-1)=√2-1, ②1/√3+√2=√解:(1)规律为:1/[√n+√(n-1)]=√n-√(n-1)(2)1/(√2+1)+1/(√3+√2+1/(√4+√3)+……+1/(

观察下列各运算:(√2-1)(√2+1)=1(√3-√2)(√3+√2)=[1/(1+√2)+1/(√2+√3)+1/(√3+√4)+……+1/(√2001+√2002)+1/(√2002+

观察下列各个等式:1/√2+1=√2-1,1/√3+√2=√3-√2,1..=(√2-1+√3-√2+√4-√3+.√2004-√2003)(√2004+1)=(√2004-1)(√2004+1)=

(√2+1)(√2-1)=1;(√3+√2)(√3-√2)=1;(√4+√3)(√观察上面的规律,计算下列式子的值:【1/(√2+1)+1/(√3+√2+1/(√4+√3)+……+1/(√2012

相关文档
sytn.net | beabigtree.com | ppcq.net | sgdd.net | ceqiong.net | 网站首页 | 网站地图
All rights reserved Powered by www.tfsf.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com