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16 tháng 11 2020
B=1+1/2*(1+2)*2/2+1/3*(1+3)*3/2+....+1/30*(1+30)*30/2
=1+1+2/2+1+3/2+...+1+30/2
=2+3+4+...+31/2
=(31+2)*30/2 phần 2
=) B=247,5
28 tháng 7 2019
a,
\(A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{900}\right)\\ =\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)...\left(1-\frac{1}{30}\right)\left(1+\frac{1}{30}\right)\\ =\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{31}{30}\\ =\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{31}{30}\\ =\frac{1\cdot2\cdot...\cdot29}{2\cdot3\cdot...\cdot30}\cdot\frac{3\cdot4\cdot...\cdot31}{2\cdot3\cdot...\cdot30}\\ =\frac{1}{30}\cdot\frac{31}{2}=\frac{31}{60}\)
b,
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+...+\frac{100-98}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\\ =\frac{1}{2}\cdot\frac{4450-1}{9900}=\frac{1}{2}\cdot\frac{4449}{9900}=\frac{4449}{19800}=\frac{1483}{6600}\)
c, (Chịu :V)
d,
\(D=\frac{1}{3}\left(\frac{3}{1\cdot2\cdot3\cdot4}+\frac{3}{2\cdot3\cdot4\cdot5}+...+\frac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{4-1}{1\cdot2\cdot3\cdot4}+\frac{5-2}{2\cdot3\cdot4\cdot5}+...+\frac{30-27}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29}-\frac{1}{28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{6}-\frac{1}{24630}\right)\\ =\frac{228}{4105}\)
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.
16 tháng 10 2021
\(1,\\ a,\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow\left[{}\begin{matrix}x-4=4\\x-4=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=0\end{matrix}\right.\\ c,\Leftrightarrow2x+1=-2\Leftrightarrow x=-\dfrac{3}{2}\\ 2,\\ a,=1\\ b,=\left(\dfrac{13}{4}\right)^2=\dfrac{169}{16}\\ c,=\left(-\dfrac{7}{4}\right)^2=\dfrac{49}{16}\\ d,=\left(\dfrac{3}{7}\right)^{20}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^8=...\\ e,=\left(3\cdot5\cdot\dfrac{2}{3}\right)^2=10^2=100\)
JN
6 tháng 8 2018
a , \(\frac{7}{8}:\frac{1}{6}+\frac{7}{8}.\frac{-7}{18}\)
= \(\frac{21}{4}+\frac{-49}{144}=\frac{707}{144}\)
b, -1 : (-5) + \(\frac{1}{15}-\frac{-1}{15}\)
= \(\frac{1}{5}+0=\frac{1}{5}\)
c, \(\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
= \(\frac{9}{10}-\frac{10-9}{10.9}-\frac{9-8}{9.8}-\frac{8-7}{8.7}-\frac{7-6}{7.6}-\frac{6-5}{6.5}-\frac{5-4}{5.4}-\frac{4-3}{4.3}-\frac{3-2}{3.2}.\frac{2-1}{2.1}\)
= \(\frac{9}{10}-1-\frac{1}{10}-1-\frac{1}{9}-1-\frac{1}{8}-1-\frac{1}{7}-1-\frac{1}{6}-1-\frac{1}{5}-1-\frac{1}{4}-1-\frac{1}{3}-1-\frac{1}{2}\)
= \(\frac{9}{10}-\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+...+\frac{1}{2}\right)\)
= \(\frac{9}{10}-9-1,928=\frac{9}{10}-7,071=-6.171\)
CK
Tìm x biết :
\(\frac{1}{1×2×3×4}+\frac{1}{2×3×4×5}+\frac{1}{3×4×5×6}+...+\frac{1}{27×28×29×30}×x=-3\)
0
RG
0
Đặt \(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+............+\frac{1}{30}.\left(1+2+3+..........+30\right)\)
\(\Rightarrow A=1+\frac{1+2}{2}+\frac{1+2+3}{3}+.........+\frac{1+2+3+.........+30}{30}\)
\(\Rightarrow A=1+\frac{\frac{\left(1+2\right).2}{2}}{2}+\frac{\frac{\left(1+3\right).3}{2}}{3}+..............+\frac{\frac{\left(1+30\right).30}{2}}{30}\)
\(\Rightarrow A=1+\frac{\left(1+2\right).2}{2.2}+\frac{\left(1+3\right).3}{2.3}+..............+\frac{\left(1+30\right).30}{2.30}\)
\(\Rightarrow A=1+\frac{1+2}{2}+\frac{1+3}{2}+............+\frac{1+30}{2}\)
\(\Rightarrow A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+.................+\frac{31}{2}\)
\(\Rightarrow A=\left(2+3+4+.............+31\right).\frac{1}{2}\)
\(\Rightarrow A=495.\frac{1}{2}=\frac{495}{2}\)
Vậy \(A=\frac{495}{2}\)
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