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nhân trước, cộng lại
quy đồng
kết quả là số tìm được
Ta có :
\(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+..............+\dfrac{1}{99.100}\)
\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...........+\dfrac{1}{99}-\dfrac{1}{100}\)
\(S=1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99x100}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}\)
=\(\frac{99}{100}\)
\(B=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2022.2023}\)
\(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\)
\(B=\dfrac{1}{2}-\dfrac{1}{2023}=\dfrac{2021}{4046}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{11\cdot12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}\)
a) \(\frac{2^{12}x3^5-4^6.9^2}{\left(2^2x3\right)^6+8^4x3^5}=\frac{2^{12}x3^5+\left(2^2\right)^6x\left(3^2\right)^2}{2^{12}x3^6+\left(2^3\right)^4x3^5}\)
\(=\frac{2^{12}x3^5-2^{12}x3^4}{2^{12}x3^6+2^{12}x3^5}=\frac{2^{12}x3^4x\left(3-1\right)}{2^{12}x3^5x\left(3+1\right)}\)
\(=\frac{2}{3.4}=\frac{1}{3.2}=\frac{1}{6}\)
b) \(\frac{1}{9x10}-\frac{1}{8x9}-\frac{1}{7x8}-\frac{1}{6x7}-\frac{1}{5x6}-\frac{1}{4x5}-\frac{1}{3x4}-\frac{1}{2x3}-\frac{1}{1x2}\)
\(=-\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+\frac{1}{9x10}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=-\left(1-\frac{1}{10}\right)\)
\(=\frac{-9}{10}\)
sorry bn nha! mk ko bk lm phần c
Ta có:
\(S = \frac{1}{2 \cdot 3} + \frac{1}{4 \cdot 5} + \frac{1}{6 \cdot 7} + . . . + \frac{1}{2022 \cdot 2023}\)
Với:
\(\frac{1}{n \left(\right. n + 1 \left.\right)} = \frac{1}{n} - \frac{1}{n + 1}\)
Suy ra:
\(S = \left(\right. \frac{1}{2} - \frac{1}{3} \left.\right) + \left(\right. \frac{1}{4} - \frac{1}{5} \left.\right) + \left(\right. \frac{1}{6} - \frac{1}{7} \left.\right) + . . . + \left(\right. \frac{1}{2022} - \frac{1}{2023} \left.\right)\)
Do đó:
\(S = \left(\right. \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + . . . + \frac{1}{2022} \left.\right) - \left(\right. \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + . . . + \frac{1}{2023} \left.\right)\)
Xét từng cặp:
\(\frac{1}{2} > \frac{1}{3} , \&\text{nbsp}; \frac{1}{4} > \frac{1}{5} , \&\text{nbsp}; . . . , \&\text{nbsp}; \frac{1}{2022} > \frac{1}{2023}\)
Suy ra:
\(S > 0\)
Mặt khác:
\(\frac{1}{n \left(\right. n + 1 \left.\right)} < \frac{1}{n^{2}} \Rightarrow S < \frac{1}{2^{2}} + \frac{1}{4^{2}} + . . . < \frac{1}{2}\)
Ta có:
\(\frac{1011}{2023} \approx 0,4997 < \frac{1}{2}\)
Và thực tế tổng \(S\) nhỏ hơn giá trị này.
Kết luận:
\(S < \frac{1011}{2023}\)