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bao quynh Cao bạn ơi hình như bn làm sai đề ạ 7/4 mà sao lại 4/7 ạ
Trả lời :
- 2 bn kia ở trong câu hỏi này có ai làm đúng đâu.
- Chúc bạn học tốt !
- Tk cho mk nha !
Đặt \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2015}-\frac{1}{2016}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{1008}\right)\)
\(A=\frac{1}{1009}+\frac{1}{1010}+.....+\frac{1}{2016}\)
Khi đó \(\frac{\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\right)}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{A}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=1\)
Bạn xem lời giải của mình nhé:
Giải:
Bài 2:
Ta xét A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1+\left(\frac{1}{2}-1\right)+\frac{1}{3}+\left(\frac{1}{4}-\frac{2}{4}\right)+...+\frac{1}{2015}+\left(\frac{1}{2016}-\frac{2}{2016}\right)\\ =1+\frac{1}{2}-1+\frac{1}{3}+\frac{1}{4}-\frac{1}{2}+...+\frac{1}{2015}+\frac{1}{2016}-\frac{1}{1008}\)
\(=\left(1-1\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{1008}-\frac{1}{1008}\right)+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(\Rightarrow\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =1\)
Chúc bạn học tốt!
E = 2/5.10 + 2/10.15 + ... + 2/35.40
E = 2/5.(1/5 - 1/10 + 1/10 - 1/15 + ... + 1/35 - 1/40)
E = 2/5.(1/5 - 1/40)
E = 2/5.7/40
E = 7/100
E = \(\frac{2}{5.10}+\frac{2}{10.15}+...+\frac{2}{35.40}\)
= \(\frac{2}{5}.\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{35.40}\right)\)
= \(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{35}-\frac{1}{40}\right)\)
= \(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{40}\right)\)
= \(\frac{2}{5}.\frac{7}{40}\)
= \(\frac{7}{100}\)
a) \(y^{2015}=y^{2020}\)
\(\Leftrightarrow y^{2020}-y^{2015}=0\)
\(\Leftrightarrow y^{2015}.\left(y^5-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y^{2015}=0\\y^5-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}y=0\\y=1\end{cases}}}\)
Vậy ...
b) \(\left(2y-1\right)^{50}=\left(2y-1\right)^1\)
\(\Leftrightarrow\left(2y-1\right)^{50}-\left(2y-1\right)^1=0\)
\(\Leftrightarrow\left(2y-1\right)^1.\left[\left(2y-1\right)^{49}-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2y-1\right)^1=0\\\left(2y-1\right)^{49}-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}y=\frac{1}{2}\\y=1\end{cases}}\)
Vậy...
\(A=\left(\frac{1}{2016}+1\right)+\left(\frac{2}{2015}+1\right)+...+\left(\frac{2015}{2}+1\right)+1\)
= \(\frac{2017}{2016}+\frac{2017}{2015}+\frac{2017}{2014}+...\frac{2017}{2}+\frac{2017}{2017}\)
= \(2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}\)
= 2017
Chúc bạn học giỏi!
\(3C=1+\frac{1}{3}+.....+\frac{1}{3^{2015}}\)
\(\Rightarrow3C-C=2C=\left(1+\frac{1}{3}+.....+\frac{1}{3^{2014}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}.....+\frac{1}{3^{2015}}\right)=1-\frac{1}{3^{2015}}\)
\(\Rightarrow C=\frac{3^{2015}-1}{3^{2015}.2}\)
\(D=4\left(1+\frac{1}{3}+....+\frac{1}{3^5}\right)\)
\(\Rightarrow3D=4\left(3+1+....+\frac{1}{3^4}\right)\)
\(\Rightarrow3D-D=2D=4\left(3+1+....+\frac{1}{3^4}\right)-4\left(1+\frac{1}{3}+....+\frac{1}{3^5}\right)\)
\(\Rightarrow2D=4\left(3-\frac{1}{3^5}\right)\Rightarrow D=2\left(3-\frac{1}{3^5}\right)\)
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