Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80} \)
\(=\frac{1}{3}.(\frac{1}{20}-\frac{1}{23})+\frac{1}{3}.(\frac{1}{23}-\frac{1}{26})+...+\frac{1}{3}.(\frac{1}{77}-\frac{1}{80})\)
=\(\frac{1}{3}.(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80})\)
=\(\frac{1}{3}.(\frac{1}{20}-\frac{1}{80})\)
=\(\frac{1}{3}.\frac{3}{80}\)
=\(\frac{1}{80}\)<\(\frac{1}{9}\)
Vậy tổng trên nhỏ hơn \(\frac{1}{9}\)
Chứng minh rằng
\(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}< \frac{1}{9}\)
Đặt A=\(\frac{1}{20.23}+\frac{1}{23.26}+....+\frac{1}{77.80}\)
=>A=\(\frac{1}{3}\).(\(\frac{3}{20.23}+\frac{3}{23.26}+....+\frac{3}{77.80}\))
=>A=\(\frac{1}{3}\).(\(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+.....+\frac{1}{77}-\frac{1}{80}\))
=>A=\(\frac{1}{3}\).(\(\frac{1}{20}-\frac{1}{80}\))
=>A=\(\frac{1}{3}.\frac{3}{80}\)
=>A=\(\frac{1}{80}\)
Do \(\frac{1}{80}\)<\(\frac{1}{9}\)
Nên \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+....+\frac{1}{77.80}< \frac{1}{9}\)
\(=\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\frac{3}{80}\)
\(=\frac{1}{80}< \frac{1}{9}\)
Ta có: \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\)
= \(\frac{1}{3.}\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)
= \(\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+....+\frac{1}{77}-\frac{1}{80}\right)\)
= \(\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
= \(\frac{1}{3}.\frac{3}{80}=\frac{1}{80}< \frac{1}{9}\)
Ta có \(\dfrac{1}{20.23}+\dfrac{1}{23.26}+...+\dfrac{1}{77.80}=\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+...+\dfrac{1}{77}-\dfrac{1}{80}\right)=\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{80}\right)=\dfrac{1}{3}.\dfrac{3}{80}=\dfrac{1}{80}< \dfrac{1}{9}\)
\(\frac{1}{20.23} +\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\)
= \(\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)
= \(\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\right)\)
= \(\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
= \(\frac{1}{3}.\frac{3}{80}\)
= \(\frac{1}{80}\) < \(\frac{1}{9}\)
⇒ \(\frac{1}{20.23} +\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\) < \(\frac{1}{9}\) (ĐPCM)
1/20.23+1/23.26+1/26.29+...+1/77.80 < 3/20.23+3/23.26+3/26.29+...+3/77.80=1/20-1/23+1/23-1/26+...+1/77-1/80
=1/20-1/80=3/80 < 1/9 = 3/27
\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\)
\(=\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{27.80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\frac{3}{80}\)
\(=\frac{1}{80}< \frac{1}{9}\)
3^2= 9
Vậy thì sẽ là:
9/ 20.23+ 9/ 23.26+...9/77.80
cách nhau 3 bỏ 3 ra ngoài
= 3(3/20.23+...3/77.80)
=3(3/20-3/23+3/23-3/26+.....+3/77-3/80)
=3(3/20-3/80)
=3. 9/80
=27/80<1
32=9
\(\frac{3^2}{20.23}\)+\(\frac{3^2}{23.26}\)+...+\(\frac{3^2}{77.80}\)
=\(\frac{9}{20.23}\)+\(\frac{9}{23.26}\)+...+\(\frac{9}{77.80}\)
=3(\(\frac{3}{20.23}\)+\(\frac{3}{23.26}\)+...+\(\frac{3}{77.80}\))
=3(\(\frac{1}{20}\)-\(\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\))
=3(\(\frac{1}{20}-\frac{1}{80}\))
=3(\(\frac{4}{80}-\frac{1}{80}\))
=3.\(\frac{3}{80}\)
=\(\frac{9}{80}\)<1
Vậy\(\frac{9}{80}< 1\)
\(\dfrac{3^2}{20.23}\)+\(\dfrac{3^2}{23.26}\)+...+\(\dfrac{3^2}{77.80}\)
=> \(\dfrac{9}{20.23}+...+\dfrac{9}{77.80}\)
= 9.\(\left(\dfrac{1}{20.23}+...+\dfrac{1}{77.80}\right)\)
\(=9.\left(\dfrac{1}{20.3}-\dfrac{1}{23.3}+\dfrac{1}{23.3}-\dfrac{1}{26.3}+...+\dfrac{1}{77.3}-\dfrac{1}{80.3}\right)\)= \(9.\left(\dfrac{1}{20.3}-\dfrac{1}{80.3}\right)\)
\(=9.\dfrac{1}{80}\)=\(\dfrac{9}{80}=0,1125< 1.\)
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 \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}< \frac{1}{9}\)