Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80} \)
\(=\frac{1}{3}.(\frac{1}{20}-\frac{1}{23})+\frac{1}{3}.(\frac{1}{23}-\frac{1}{26})+...+\frac{1}{3}.(\frac{1}{77}-\frac{1}{80})\)
=\(\frac{1}{3}.(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80})\)
=\(\frac{1}{3}.(\frac{1}{20}-\frac{1}{80})\)
=\(\frac{1}{3}.\frac{3}{80}\)
=\(\frac{1}{80}\)<\(\frac{1}{9}\)
Vậy tổng trên nhỏ hơn \(\frac{1}{9}\)
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 \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}< \frac{1}{9}\)
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)
4 + 4^3 + 4^5 + 4^7 + ... + 4^23
= ( 4 + 4^3 ) + ( 4^5 + 4^7 ) +.....+ ( 4^22 + 4^23)
=4( 1+16 ) + 4^5( 1+16 ) +....+ 4^22( 1+ 16 )
=4 x 17 + 4^5 x 17+....+ 4^22 x 17 chia hết cho 68
Câu 2:
1+3+3^2+3^3+....+3^2000
=( 1+3 +3^2 ) + ( 3^3 + 3^4 + 3^5 ) +.....+ ( 3^ 1998 + 3^1999 + 3^2000)
=1( 1+ 3 + 9 ) + 3^3 + ( 1+ 3 + 9 ) +......+ 3^1998+( 1+ 3 + 9 )
= 1 x 13+ 3^3 x 13 +......+ 3^1998 x 13 chia hết cho 13
k mk nha lần sau mk k lại
Câu 1 nha : 4+4^3+4^5+4^7+....+4^23 = (4+4^3)+(4^5+4^7)+....+(4^21+4^23)
= 68 + 4^4.(4+4^3)+....+4^20.(4+4^3) = 68 + 4^4.68 + .... + 4^20.68
=68.(1+4^4+....+4^20) chia hết cho 68
Câu 2 nha 1+3+3^2+...+3^2000 = (1+3+3^2)+(3^3+3^4+3^5)+....+(3^1998+3^1999+3^2000)
= 13 + 3^3.(1+3+3^2)+....+3^1998.(1+3+3^2) = 13+3^3.13+....+3^1998.13
=13.(1+3^3+....+3^1998) chia hết cho 13
\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\)
\(=\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{27.80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\frac{3}{80}\)
\(=\frac{1}{80}< \frac{1}{9}\)
ko bt sory nha
Ta có:
\(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\)
\(=\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\frac{3}{80}\)
\(=\frac{1}{80}\)
Mà \(\frac{1}{80}< \frac{1}{9}\)
\(\Rightarrow\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}< \frac{1}{9}\left(đpcm\right)\)
Học tốt