Sửa đề bài `2a)(11/12+11/(12.23)+11/(23.24)+...+11/(89.100))+x=2/3`
`→a)(11/12+11/(12.23)+11/(23.34)+...+11/(89.100))+x=2/3`
Bài 1:
`c)1/(1.3)+1/(3.5)+1/(5.7)+...+1/(99.101)`
`=1/2 .(2/(1.3)+2/(3.5)+2/(5.7)+...+2/(99.101))`
`=1/2 .(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)`
`=1/2 .(1-1/101)`
`=1/2 .(101/101-1/101)`
`=1/2 . 100/101`
`=50/101`
Bài 2:
`a)(11/12+11/(12.23)+11/(23.34)+...+11/(89.100))+x=2/3`
`(11/12+1/12-1/23+1/23-1/34+...+1/89-1/100)+x=2/3`
`(11/12+1/12-1/100)+x=2/3`
`(1-1/100)+x=2/3`
`(100/100-1/100)+x=2/3`
`99/100+x=2/3`
`x=2/3-99/100`
`x=200/300-297/300`
`x=-97/300`
`b)(2/(11.13)+2/(13.15)+...+2/(19.21))-x+221/231=4/3`
`(1/11-1/13+1/13-1/15+...+1/19-1/21)+221/231-x=4/3`
`(1/11-1/21)+221/231-x=4/3`
`(21/231-11/231)+221/231-x=4/3`
`10/231+221/231-x=4/3`
`231/231-x=4/3`
`1-x=4/3`
`x=1-4/3`
`x=3/3-4/3`
`x=-1/3`
Bài 3:
`1/(1.2)+1/(2.3)+1/(3.4)+...+1/(49.50)`
`=1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50`
`=1-1/50`
`=50/50-1/50`
`=49/50`
Vì `49/50<1`
`⇒1/(1.2)+1/(2.3)+1/(3.4)+...+1/(49.50)<1(đpcm)`
