CHUNG MINH:1/3 26/07/2021 Bởi Sadie CHUNG MINH:1/3 { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " CHUNG MINH:1/3
Đáp án: `+)` Ta có : `1/3^2 > 1/3.4` `1/4^2 > 1/4.5` `…` `1/100^2 > 1/100.101` `to 1/3^2+1/4^2+…+1/100^2>1/3.4+1/4.5+…+1/100.101` `to A>1/3-1/4+1/4-1/5+…+1/100-1/101>1/3 \ \ \ (1)` `+)` Ta có : `1/3^2 < 1/2.3` `1/4^2 < 1/3.4` `…` `1/100^2 < 1/99.100` `to 1/3^2+1/4^2+…+1/100^2<1/2.3+1/3.4+…+1/99.100` `to A<1/2-1/3+1/3-1/4+…+1/99-1/100` `to A<1/2-1/100<1/2 \ \ \ (2)` Từ `(1)` và `(2)“=> 1/3<A<1/2` Bình luận
Đáp án: `A=1/3^2+1/4^2+1/5^2+…+1/100^2` `=>A=1/3.3+1/4.4+1/5.5+…+1/100.100` `=>1/3.4+1/4.5+1/5.6+…+1/100.101<A<1/2.3+1/3.4+1/4.5+…+1/99.100` `=>1/3-1/4+1/4-1/5+1/5-1/6+…+1/100-1/101<A<1/2-1/3+1/3-1/4+1/4-1/5+…+1/99-1/100` `=>1/3-1/101<A<1/2-1/100` `=>1/3<A<1/2` Vậy `1/3<A<1/2`. Bình luận
Đáp án:
`+)`
Ta có :
`1/3^2 > 1/3.4`
`1/4^2 > 1/4.5`
`…`
`1/100^2 > 1/100.101`
`to 1/3^2+1/4^2+…+1/100^2>1/3.4+1/4.5+…+1/100.101`
`to A>1/3-1/4+1/4-1/5+…+1/100-1/101>1/3 \ \ \ (1)`
`+)`
Ta có :
`1/3^2 < 1/2.3`
`1/4^2 < 1/3.4`
`…`
`1/100^2 < 1/99.100`
`to 1/3^2+1/4^2+…+1/100^2<1/2.3+1/3.4+…+1/99.100`
`to A<1/2-1/3+1/3-1/4+…+1/99-1/100`
`to A<1/2-1/100<1/2 \ \ \ (2)`
Từ `(1)` và `(2)“=> 1/3<A<1/2`
Đáp án:
`A=1/3^2+1/4^2+1/5^2+…+1/100^2`
`=>A=1/3.3+1/4.4+1/5.5+…+1/100.100`
`=>1/3.4+1/4.5+1/5.6+…+1/100.101<A<1/2.3+1/3.4+1/4.5+…+1/99.100`
`=>1/3-1/4+1/4-1/5+1/5-1/6+…+1/100-1/101<A<1/2-1/3+1/3-1/4+1/4-1/5+…+1/99-1/100`
`=>1/3-1/101<A<1/2-1/100`
`=>1/3<A<1/2`
Vậy `1/3<A<1/2`.