Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
đặt \(A=\frac{3^4}{20\cdot23}+\frac{3^4}{23\cdot26}+...+\frac{3^4}{77\cdot80}\)
\(A=3^3\left(\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+...+\frac{3}{77\cdot80}\right)\)
\(A=3^3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(A=3^3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(A=3^3\cdot\frac{3}{80}\)
\(A=\frac{3^4}{80}=\frac{81}{80}>1\)
\(\frac{3^4}{20.23}+\frac{3^4}{23.26}+...+\frac{3^4}{77.80}\)
\(=3^3\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)
\(=3^3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=3^3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{3^3.3}{80}\)
\(=\frac{3^4}{80}\)
\(=\frac{81}{80}\)
\(=\frac{80}{80}+\frac{1}{80}\)
\(=1+\frac{1}{80}\)
=> Biểu thức trên lớn hơn 1
Đặt vế trái là B
\(3B=\frac{23-20}{20.23}+\frac{26-23}{23.26}+\frac{29-26}{26.29}+...+\frac{80-77}{77.80}\)
\(3B=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}=\frac{1}{20}-\frac{1}{80}\)
\(3B=\frac{3}{80}\Rightarrow B=\frac{1}{80}< \frac{1}{9}\)
Ta có: \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\)
\(=\frac{1}{3}\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)
\(=\frac{1}{3}\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\frac{3}{80}=\frac{1}{80}< \frac{1}{9}\)
Vậy \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}< \frac{1}{9}\)
Ta có \(\dfrac{1}{20.23}+\dfrac{1}{23.26}+...+\dfrac{1}{77.80}=\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+...+\dfrac{1}{77}-\dfrac{1}{80}\right)=\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{80}\right)=\dfrac{1}{3}.\dfrac{3}{80}=\dfrac{1}{80}< \dfrac{1}{9}\)
Ta có : \(\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}=3\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)=3\left(\frac{1}{20}-\frac{1}{80}\right)=3.\frac{3}{80}=\frac{9}{80}< 1\)
Bài làm:
Mình nghĩ đề sai rồi, phải như vậy nè:
\(A=\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\)
\(A=\frac{23-20}{20.23}+\frac{26-23}{23.26}+...+\frac{80-77}{77.80}\)
\(A=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\)
\(A=\frac{1}{20}-\frac{1}{80}\)
\(A=\frac{3}{80}\)
Vậy \(A=\frac{3}{80}\)
\(A=\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\)
\(\Rightarrow A=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\)
\(\Rightarrow A=\frac{1}{20}-\frac{1}{80}\)
\(\Rightarrow A=\frac{3}{80}\)
\(\frac{1}{20.23} +\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\)
= \(\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)
= \(\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\right)\)
= \(\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
= \(\frac{1}{3}.\frac{3}{80}\)
= \(\frac{1}{80}\) < \(\frac{1}{9}\)
⇒ \(\frac{1}{20.23} +\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\) < \(\frac{1}{9}\) (ĐPCM)
\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80} \)
\(=\frac{1}{3}.(\frac{1}{20}-\frac{1}{23})+\frac{1}{3}.(\frac{1}{23}-\frac{1}{26})+...+\frac{1}{3}.(\frac{1}{77}-\frac{1}{80})\)
=\(\frac{1}{3}.(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80})\)
=\(\frac{1}{3}.(\frac{1}{20}-\frac{1}{80})\)
=\(\frac{1}{3}.\frac{3}{80}\)
=\(\frac{1}{80}\)<\(\frac{1}{9}\)
Vậy tổng trên nhỏ hơn \(\frac{1}{9}\)
\(\frac{A}{3}=\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\)
\(\frac{A}{3}=\frac{23-20}{20.23}+\frac{26-23}{23.26}+...+\frac{80-77}{77.80}\)
\(\frac{A}{3}=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}=\frac{1}{20}-\frac{1}{80}=\frac{3}{80}\Rightarrow A=\frac{9}{80}< 1\)