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\(=\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\frac{3}{80}\)
\(=\frac{1}{80}< \frac{1}{9}\)
Ta có: \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\)
= \(\frac{1}{3.}\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)
= \(\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+....+\frac{1}{77}-\frac{1}{80}\right)\)
= \(\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
= \(\frac{1}{3}.\frac{3}{80}=\frac{1}{80}< \frac{1}{9}\)
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 \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+....+\frac{1}{77.80}< \frac{1}{9}\)
\(3\left(\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+....+\frac{3}{77\cdot80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+.....+\frac{1}{77}-\frac{1}{80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{3}{20}-\frac{3}{80}\)
\(< 1\)
A=\(3 \left(\right. \frac{3}{20.23} + \frac{3}{23.26} + \frac{3}{26.29} + . . . + \frac{3}{77.80} \left.\right)\)
A\(= 3 \left(\right. \frac{1}{20} - \frac{1}{23} + \frac{1}{23} - \frac{1}{26} + \frac{1}{26} - \frac{1}{29} + . . . + \frac{1}{77} - \frac{1}{80} \left.\right)\)
\(A = 3 \left(\right. \frac{1}{20} - \frac{1}{80} \left.\right)\)
A\(= 3 \left(\right. \frac{4}{80} - \frac{1}{80} \left.\right)\)
A\(= 3. \frac{3}{80}\)
A\(= \frac{9}{80}\)
\(\frac{9}{80} < 1\)
⇒A<1
Ta có B= 3(\(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\) )
B= 3\(\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\right)\)
B= 3.(\(\frac{1}{20}-\frac{1}{80}\))
B=3.\(\frac{3}{80}\)=\(\frac{9}{80}\)
\(\frac{3}{9}B=\frac{3}{9}.\left(\frac{9}{20.23}+\frac{9}{23.26}+\frac{9}{26.29}+...+\frac{9}{77.80}\right)\)
=> \(\frac{3}{9}B=\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+....+\frac{3}{77.80}\)
=\(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+....+\frac{1}{77}-\frac{1}{80}\)
=\(\frac{1}{20}-\frac{1}{80}=\frac{3}{80}\)
=> \(B=\frac{3}{80}:\frac{3}{9}=\frac{3}{80}.\frac{9}{3}=\frac{9}{80}\)
=\(3\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\right)\)\(=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=3\left(\frac{4}{80}-\frac{1}{80}\right)\)
\(=3.\frac{3}{80}\)
\(=\frac{9}{80}\)
\(A=\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}\)
\(\frac{A}{3}=\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\)
\(\frac{A}{3}=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\)
\(\frac{A}{3}=\frac{1}{20}-\frac{1}{80}\)
\(\frac{A}{3}=\frac{3}{80}\)
\(A=\frac{3}{80}.3=\frac{9}{80}< 1\)
Đặt A=32/20.23+32/23.26+....................+32/77.80
A=3(3/20.23+3/23.26+.........+3/77.80)
A=3(1/20-1/23+1/23-1/26+.+1/77-1/80)
A=3(1/20-1/80)
A=3.3/80
A=9/80 Mà A=9/80<1 =>A<1 (đpcm)
Ta có
\(A=\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}\)
\(A=3^2\left(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\right)\)
\(A=3^2\cdot\frac{1}{3}\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(A=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(A=3\cdot\frac{3}{80}=\frac{9}{80}< 1\left(9< 80\right)\)
A=\(3 \left(\right. \frac{3}{20.23} + \frac{3}{23.26} + \frac{3}{26.29} + . . . + \frac{3}{77.80} \left.\right)\)
A\(= 3 \left(\right. \frac{1}{20} - \frac{1}{23} + \frac{1}{23} - \frac{1}{26} + \frac{1}{26} - \frac{1}{29} + . . . + \frac{1}{77} - \frac{1}{80} \left.\right)\)
\(A = 3 \left(\right. \frac{1}{20} - \frac{1}{80} \left.\right)\)
A\(= 3 \left(\right. \frac{4}{80} - \frac{1}{80} \left.\right)\)
A\(= 3. \frac{3}{80}\)
A\(= \frac{9}{80}\)
\(\frac{9}{80} < 1\)
⇒A<1
Ta có : \(\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}=3\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)=3\left(\frac{1}{20}-\frac{1}{80}\right)=3.\frac{3}{80}=\frac{9}{80}< 1\)
\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\)
\(=\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{27.80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\frac{3}{80}\)
\(=\frac{1}{80}< \frac{1}{9}\)
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