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\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{3}-\dfrac{1}{20}=\dfrac{17}{60}< \dfrac{1}{2}\)
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{19.20}< \dfrac{1}{2}\)
=> \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{20}< \dfrac{1}{2}\)
=> \(\dfrac{1}{3}-\dfrac{1}{20}< \dfrac{1}{2}\)
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\(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\cdot\cdot\cdot+\dfrac{1}{18\cdot19}+\dfrac{1}{19\cdot20}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\cdot\cdot\cdot+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=\dfrac{1}{2}-\dfrac{1}{20}\)
\(=\dfrac{9}{20}\)
#\(Urushi\)☕
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Lời giải:
Ta có:
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{100-99}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
Vậy ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{19\cdot20}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1-\dfrac{1}{20}=\dfrac{19}{20}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+....+\dfrac{1}{19\cdot20}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{20}\)
\(A=1-\dfrac{1}{20}\)
\(A=\dfrac{19}{20}\)
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1/2.3 + 1/3.4 + 1/4.5 + ... + 1/19.20
= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/19 - 1/20
= 1/2 - 1/20
= 9/20
k đii
1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/19 - 1/20
1/2 - 1/20
9/20
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.3-2/2.3 + 4-3/3.4 + 5-4/4.5 + 6-5/5.6 +...+ 20-19/19.20=18/x
1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 +...+ 1/19 - 1/20=18/x
1/2 - 1/20=18/x
10/20 - 1/20=18/x
9/20=18/x
18/40=18/x
=>x=40
Vậy x=40
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(b)\) Đặt \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\) ta có :
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}-0=\frac{1}{2}\)
\(\Rightarrow\)\(A< \frac{1}{2}\) ( đpcm )
Vậy \(A< \frac{1}{2}\)
Chúc bạn học tốt ~
\(a)\frac{9.25-63}{3.30+153}\)
\(=\frac{9.25-9.7}{3.30+3.51}\)
\(=\frac{9.\left(25-7\right)}{3.\left(30+51\right)}\)
\(=\frac{9.18}{3.81}\)
\(=\frac{1.6}{1.9}\)
\(=\frac{6}{9}\)
\(=\frac{2}{3}\)
b ) \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(Đpcm\right)\)
Chúc bạn học tốt !!!
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\(\frac{1}{n\left(n+1\right)}=\frac{\left(n+1\right)-n}{n\left(n+1\right)}=\frac{\left(n+1\right)}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
\(\Rightarrow\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
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\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)
=1-1/20
=19/20
1 3.4 + 1 4.5 + ... + 1 19.20 = 1 3 − 1 20 = 17 60 < 1 2