Cho A= 1/12+1/13+1/14+…+1/100. So sánh A với 1 01/08/2021 Bởi Daisy Cho A= 1/12+1/13+1/14+…+1/100. So sánh A với 1
Đáp án: `A = 1/12 + 1/13 + 1/14 + …. + 1/100` `⇔ A = (1/12 + 1/13 + 1/14 + 1/15) + (1/100 + 1/100 + 1/100 +…. + 1/100)` `⇔ A = 543/1820 + (85 . 1/100)` `⇔ A= 543/1820 + 17/20` `⇔ A = 209/182` Ta có : `209/182 = 1,1 > 1` `-> 209/182 > 1` `-> A > 1` Bình luận
Giải thích các bước giải: Ta có: $\begin{array}{l}A = \dfrac{1}{{12}} + \dfrac{1}{{13}} + \dfrac{1}{{14}} + … + \dfrac{1}{{100}}\\ \Rightarrow A = \left( {\dfrac{1}{{12}} + \dfrac{1}{{13}} + \dfrac{1}{{14}} + \dfrac{1}{{15}}} \right) + \left( {\dfrac{1}{{16}} + \dfrac{1}{{17}}… + \dfrac{1}{{100}}} \right)\\ \Rightarrow A > \left( {\dfrac{1}{{12}} + \dfrac{1}{{13}} + \dfrac{1}{{14}} + \dfrac{1}{{15}}} \right) + \left( {\dfrac{1}{{100}} + \dfrac{1}{{100}} + … + \dfrac{1}{{100}}} \right)\\\left( {\text{Có 85 phân số }\dfrac{1}{100}} \right)\\ \Rightarrow A > \dfrac{1}{{12}} + \dfrac{1}{{13}} + \dfrac{1}{{14}} + \dfrac{1}{{15}} + \dfrac{{85}}{{100}}\\ \Rightarrow A > \dfrac{{209}}{{182}} > 1\\ \Rightarrow A > 1\end{array}$ Bình luận
Đáp án:
`A = 1/12 + 1/13 + 1/14 + …. + 1/100`
`⇔ A = (1/12 + 1/13 + 1/14 + 1/15) + (1/100 + 1/100 + 1/100 +…. + 1/100)`
`⇔ A = 543/1820 + (85 . 1/100)`
`⇔ A= 543/1820 + 17/20`
`⇔ A = 209/182`
Ta có : `209/182 = 1,1 > 1`
`-> 209/182 > 1`
`-> A > 1`
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
A = \dfrac{1}{{12}} + \dfrac{1}{{13}} + \dfrac{1}{{14}} + … + \dfrac{1}{{100}}\\
\Rightarrow A = \left( {\dfrac{1}{{12}} + \dfrac{1}{{13}} + \dfrac{1}{{14}} + \dfrac{1}{{15}}} \right) + \left( {\dfrac{1}{{16}} + \dfrac{1}{{17}}… + \dfrac{1}{{100}}} \right)\\
\Rightarrow A > \left( {\dfrac{1}{{12}} + \dfrac{1}{{13}} + \dfrac{1}{{14}} + \dfrac{1}{{15}}} \right) + \left( {\dfrac{1}{{100}} + \dfrac{1}{{100}} + … + \dfrac{1}{{100}}} \right)\\
\left( {\text{Có 85 phân số }\dfrac{1}{100}} \right)\\
\Rightarrow A > \dfrac{1}{{12}} + \dfrac{1}{{13}} + \dfrac{1}{{14}} + \dfrac{1}{{15}} + \dfrac{{85}}{{100}}\\
\Rightarrow A > \dfrac{{209}}{{182}} > 1\\
\Rightarrow A > 1
\end{array}$