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\(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}\)
\(=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}\cdot\frac{10}{39}=\frac{5}{39}\)
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{1.13}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}.\frac{10}{39}=\frac{5}{39}\)
Gần đúng được không
\(A=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)cách làm \(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\) tách hết ra bạn thấy cái giữa tự triệt tiêu nhau
\(B=\frac{5}{1.3}+...+\frac{5}{23.25}\) { nếu đúng là \(\frac{5}{23.5}\) thì làm đến \(\frac{5}{21.23}\) rồi cộng lẻ cái cuối
\(\frac{2B}{5}=\frac{2}{1.3}+...+\frac{1}{23.25}=1-\frac{1}{25}\) cách làm giống (a)
\(\frac{2}{5}B=1-\frac{1}{25}=\frac{24}{25}\Rightarrow B=\frac{24}{25}.\frac{5}{2}=\frac{12}{5}\)
A=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
=\(\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{19}-\frac{1}{20}\right)\)
=\(\frac{1}{2}-\frac{1}{20}=\frac{10-1}{20}=\frac{9}{20}\)
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
Mình ko bít có đúng ko nên sai đừng trách mình nhé !
\(A=\frac{7^{2011}+1}{7^{2013}+1}\)
\(7^2.A=\frac{7^{2013}+49}{7^{2013}+1}=\frac{7^{2013}+1+48}{7^{2013}+1}=\)\(\frac{7^{2013}+1}{7^{2013}+1}+\frac{48}{7^{2013}+1}=1\frac{48}{7^{2013}+1}\)
\(B=\frac{7^{2013}+1}{7^{2015}+1}\)
\(7^2.B=\)\(=\frac{7^{2015}+49}{7^{2015}+1}=\)\(\frac{7^{2015}+1+48}{7^{2015}+1}=\)\(\frac{7^{2015}+1}{7^{2015}+1}+\frac{48}{7^{2015}+1}=1\frac{48}{7^{2015}+1}\)
\(Vì\) \(1\frac{48}{7^{2013}+1}>1\frac{48}{7^{2013}+1}\)\(\Rightarrow7^2.A>7^2.B\)\(\Rightarrow A>B\)
\(Vậy\) \(A>B\)
Bài 2 nè
ta xét B trước:
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..\)\(.....+\frac{1}{2015}-\frac{1}{2016}\)
=\(\left(\frac{1}{1}+\frac{1}{3}+....+\frac{1}{2015}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{2016}\right)\)
\(=\)\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}\right)-\)\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)
vậy A:B\(=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)\(:\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)
\(=1\)
Phần a làm như sau :
x/2=1/3+1/5
x/2=8/15
x*15/30=16/30
x*15=16
x=16:15
x=16/15
Phần b làm như sau:
x=31/30
Còn phần c như sau:
x/3=16/63
x*21/63=16/63
x*21=16
x=16/21
Phần d như sau:
x/7=11/24
x*21/168=77/168
x*21=77
x=77/21
x=11/3
Đặt \(B=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2014^2}\)
Ta có : \(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(\frac{1}{5^2}< \frac{1}{4.5}\)
...
\(\frac{1}{2014^2}< \frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(\Rightarrow B< \frac{1}{2}-\frac{1}{2014}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2}=\frac{3}{4}\)
Vậy A<\(\frac{3}{4}\)
A<\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)=\(\frac{2013}{2014}\)<\(\frac{3}{4}\)
b) \(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot...\cdot\frac{100^2}{100\cdot101}=\frac{\left(1\cdot2\cdot3\cdot...\cdot100\right)}{1\cdot2\cdot3\cdot4\cdot...\cdot100}\cdot\frac{\left(1\cdot2\cdot3\cdot...\cdot100\right)}{2\cdot3\cdot4\cdot...\cdot101}=1\cdot\frac{1}{101}=\frac{1}{101}\)
a không biết
Đặt A = 1/5x10 + 1/10x15 + 1/15x20 + 1/20x25 + ... + 1/95x100
A x 5 = 5/5x10 + 5/10x15 + 5/15x20 + 5/20x25 + ... + 5/95x100
A x 5 = 1/5 - 1/10 + 1/10 - 1/15 + 1/15 - 1/20 + 1/20 - 1/25 + ... + 1/95 - 1/100
A x 5 = 1/5 - 1/100
A x 5 = 19/100
A = 19/100 : 5
A = 19/100 x 1/5
A = 19/500
Vậy A= 19/500
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