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Sửa đề: \(D=\dfrac{3}{2\cdot6}+\dfrac{3}{6\cdot10}+\dfrac{3}{10\cdot14}+...+\dfrac{3}{26\cdot30}\)
Ta có: \(D=\dfrac{3}{2\cdot6}+\dfrac{3}{6\cdot10}+\dfrac{3}{10\cdot14}+...+\dfrac{3}{26\cdot30}\)
\(=\dfrac{3}{4}\left(\dfrac{4}{2\cdot6}+\dfrac{4}{6\cdot10}+\dfrac{4}{10\cdot14}+...+\dfrac{4}{26\cdot30}\right)\)
\(=\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{26}-\dfrac{1}{30}\right)\)
\(=\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{30}\right)\)
\(=\dfrac{3}{4}\cdot\dfrac{28}{60}\)
\(=\dfrac{21}{60}=\dfrac{7}{20}\)
\(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2015.2018}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
Vậy \(A=\frac{504}{1009}.\)
\(B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)
\(=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)
\(=\frac{1}{2}-\frac{1}{106}=\frac{26}{53}\)
Vậy \(B=\frac{26}{53}.\)
Bài làm:
a) \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2015.2018}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2018}\)
\(A=\frac{1}{2}-\frac{1}{2018}\)
\(A=\frac{504}{1009}\)
b) \(B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)
\(B=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)
\(B=\frac{1}{2}-\frac{1}{106}\)
\(B=\frac{26}{53}\)
\(\dfrac{3}{2.6}\) + \(\dfrac{3}{6.10}\) + \(\dfrac{3}{10.14}\)
= \(\dfrac{3}{4}\).(\(\dfrac{4}{2.6}\) + \(\dfrac{4}{6.10}\) + \(\dfrac{4}{10.14}\))
= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}-\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{14}\))
= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{14}\))
= \(\dfrac{3}{4}\). \(\dfrac{3}{7}\)
= \(\dfrac{9}{28}\)
B = \(\dfrac{4}{1.3.5}\) + \(\dfrac{4}{3.5.7}\) + \(\dfrac{4}{5.7.9}\)
B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{3.5}\) + \(\dfrac{1}{3.5}\) - \(\dfrac{1}{5.7}\) + \(\dfrac{1}{5.7}\) - \(\dfrac{1}{7.9}\)
B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{7.9}\)
B = \(\dfrac{1}{3}\) - \(\dfrac{1}{63}\)
B = \(\dfrac{20}{63}\)
4. ( 3x+3 + 3x+1 ) = 3240
3x+3 + 3x+1 = 810
3x . 33 + 3x . 3= 810
3x. 30=810
3x = 27
3x = 33
x=3
vậy x =3
4(3𝑥+3+3𝑥+1)=3240
4(3x+{\color{#c92786}{3}}+3x+{\color{#c92786}{1}})=32404(3x+3+3x+1)=3240
4(3𝑥+4+3𝑥)=3240
Đáp án
𝑥=403/3
ta có: \(S=1+1\times2+2\times3+3\times4+...+38\times39+39\times40+40\)
\(\Rightarrow3S=1\times3+1\times2\times3+2\times3\times3+...+39\times40\times3+40\times3\)
\(3S=3+1\times2\times\left(3-0\right)+2\times3\times\left(4-1\right)+...+39\times40\times\left(41-38\right)+120\)
\(3S=3+1\times2\times3+2\times3\times4-1\times2\times3+...+39\times40\times41-38\times39\times40+120\)
\(3S=\left(3+1.2.3+...+39.40.41+120\right)-\left(1.2.3+...+38.38.40\right)\)
\(3S=3+39.40.41+120\)
\(\Rightarrow S=\left(3+39.40.41+120\right):3\)
\(S=21361\)
Đặt A = 1/2.6 + 1/6.10 + 1/10.14 + ..... + 1/102.106
=> 4A = 4/2.6 + 4/6.10 + 4/10.14 + ..... + 4/102.106
=> 4A = 1/2 - 1/6 + 1/6 - 1/10 + 1/10 - 1/14 + ... + 1/102 - 1/106
=> 4A = 1/2 - 1/106
=> 4A = 26/53
=> A = 13/106
~Study well~
#QASJ
\(\frac{1}{2.6}+\frac{1}{6.10}+...+\frac{1}{102.106}\)
\(=\frac{1}{4}.\left(\frac{4}{2.6}+\frac{4}{6.10}+...+\frac{4}{102.106}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{102}-\frac{1}{106}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{106}\right)\)
\(=\frac{1}{4}.\frac{26}{53}\)
\(=\frac{13}{106}\)
Đề sai bạn nhé.
Ta có: \(\dfrac{3}{2\cdot6}+\dfrac{3}{6\cdot10}+\dfrac{3}{10\cdot14}+...+\dfrac{3}{26\cdot30}\)
\(=\dfrac{3}{4}\left(\dfrac{4}{2\cdot6}+\dfrac{4}{6\cdot10}+\dfrac{4}{10\cdot14}+...+\dfrac{4}{26\cdot30}\right)\)
\(=\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{26}-\dfrac{1}{30}\right)\)
\(=\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{30}\right)\)
\(=\dfrac{3}{4}\cdot\dfrac{7}{15}=\dfrac{21}{60}=\dfrac{7}{20}\)