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TG
5
28 tháng 7 2015
\(A=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}=\frac{\left(2.3...29.30\right).\left(3.4.5...29.31\right)}{\left(2.3...29.30\right).\left(2.3.4...29.30\right)}=\frac{31}{2.30}=\frac{31}{60}\)
DA
0
QU
1
3 tháng 10 2018
Đặt \(A=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}\)
\(\Rightarrow A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}\)
\(\Rightarrow A=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}\)
\(\Rightarrow A=\frac{\left(1.2.3...29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}\)
\(\Rightarrow A=\frac{1}{30}.\frac{31}{2}=\frac{31}{60}\)
Vậy \(A=\frac{31}{60}\)
NT
2
LT
1
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,
\(D=\frac{1}{3}\left(\frac{3}{1\cdot2\cdot3\cdot4}+\frac{3}{2\cdot3\cdot4\cdot5}+...+\frac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{4-1}{1\cdot2\cdot3\cdot4}+\frac{5-2}{2\cdot3\cdot4\cdot5}+...+\frac{30-27}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29}-\frac{1}{28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{6}-\frac{1}{24630}\right)\\ =\frac{228}{4105}\)
Chúc bạn học tốt nha
.
GR
2 tháng 11 2017
\(a,\frac{-8}{15}.\left(-30\right).\frac{15}{-8}.\frac{9}{10}\)
\(=-\left(\frac{8}{15}.\frac{15}{8}\right).\left(30.\frac{9}{10}\right)\)
\(=-1.27
=-27\)
\(b,2\frac{1}{18}.\frac{23}{24}.\frac{9}{37}.\frac{48}{-15}\)
\(=\frac{-37.23.9.48}{18.24.37.15}=\frac{23}{15}\)
c, chịu rồi
\(Q=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.......................\dfrac{899}{30^2}\)
\(\Leftrightarrow Q=\dfrac{1.3.2.4.3.5........29.31}{2.2.3.3......30.30}\)
\(\Leftrightarrow Q=\dfrac{\left(2.3.....29.30\right)\left(3.4.5....29.31\right)}{\left(2.3.4....30\right)\left(2.3.4....30\right)}\)
\(\Leftrightarrow Q=\dfrac{31}{2.30}\)
\(\Leftrightarrow Q=\dfrac{31}{60}\)
A=\(\dfrac{3}{2^2}\)x\(\dfrac{8}{3^2}\)x\(\dfrac{15}{4^2}\)......\(\dfrac{899}{30^2}\)=\(\dfrac{1.3.2.4.3.5....29.31}{2.2.3.3.4.4....30.30}\)= (\(\dfrac{2.3....29.30}{2.3....20.30}\)).(\(\dfrac{3.4.5....29.31}{2.3.4....29.30}\))=\(\dfrac{31}{2.30}\)=\(\dfrac{31}{60}\)