3x - (1/1x2+1/2x3+.....+1/99x100)=1/1x2x3+1/2x3x4+......+1/18x19x20
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F = 1- 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2006.2007}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2006}-\frac{1}{2007}\)
\(=1-\frac{1}{2007}\)
\(=\frac{2006}{2007}\)
Ta có:
\(A=\frac{1}{1\text{x}2\text{x}3}+\frac{1}{2\text{x}3\text{x}4}+\frac{1}{3\text{x}4\text{x}5}+...+\frac{1}{18\text{x}19\text{x}20}< \frac{1}{4}\)
\(A=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{20}< \frac{1}{4}\)
\(A=1+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\frac{1}{20}< \frac{1}{4}\)
\(A=1+\frac{1}{20}< \frac{1}{4}\)
\(A=\frac{19}{20}< \frac{1}{4}\)
\(A=\frac{19}{20}< \frac{5}{20}\)
\(A>\frac{1}{4}\)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)=\frac{1}{4}-\frac{1}{760}< \frac{1}{4}\)(ĐPCM)
\(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)
=1/2-1/380
=190/380-1/380
=189/380
Gọi biểu thức trên là S. Ta có :
\(S=\dfrac{1}{1\times2\times3}+\dfrac{1}{2\times3\times4}+\dfrac{1}{3\times4\times5}+...+\dfrac{1}{18\times19\times20}\)
\(=\dfrac{1}{2}\times\left(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{18\times19\times20}\right)\)
Trước tiên, ta áp dụng : \(\dfrac{2}{a\left(a+1\right)\left(a+2\right)}=\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}\)
Ta sẽ có :
\(S=\dfrac{1}{2}\times\left(\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+...+\dfrac{1}{18\times19}-\dfrac{1}{19\times20}\right)\)
\(=\dfrac{1}{2}\times\left(\dfrac{1}{1\times2}-\dfrac{1}{19\times20}\right)\)
\(=\dfrac{1}{2}\times\dfrac{1}{1\times2}-\dfrac{1}{2}\times\dfrac{1}{19\times20}\)
\(=\dfrac{1}{4}-\dfrac{1}{760}=\dfrac{189}{760}\)
\(\Leftrightarrow3x-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\right)=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\right)\)
\(\Leftrightarrow3x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+....+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(\Leftrightarrow3x-\left(1-\frac{1}{100}\right)=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)
\(\Leftrightarrow3x-\frac{99}{100}=\frac{1}{2}\cdot\frac{189}{380}\)
\(\Leftrightarrow3x-\frac{99}{100}=\frac{189}{760}\)
\(\Leftrightarrow3x=\frac{189}{760}+\frac{99}{100}=\frac{4707}{3800}\)
\(\Leftrightarrow x=\frac{1569}{3800}\)
\(\text{Vậy }x=\frac{1569}{3800}\)
Học sinh gương mẫu của lớp thầy Phú là đây