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15 tháng 1

\(C=\left(1-\frac13\right)\left(1-\frac16\right)\left(1-\frac{1}{10}\right)\ldots\left(1-\frac{1}{210}\right)\)

\(C=\frac23\cdot\frac56\cdot\frac{9}{10}\cdot\ldots\cdot\frac{209}{210}=\frac46\cdot\frac{10}{12}\cdot\frac{18}{20}\cdot\ldots\cdot\frac{418}{420}\)

\(C=\frac{1\cdot4}{2\cdot3}\cdot\frac{2\cdot5}{3\cdot4}\cdot\frac{3\cdot6}{4\cdot5}\cdot\ldots\cdot\frac{19\cdot23}{20\cdot21}\)

\(C=\frac{\left(1\cdot2\cdot3\cdot\ldots\cdot19\right)\left(4\cdot5\cdot6\cdot\ldots\cdot23\right)}{\left(2\cdot3\cdot4\cdot\ldots\cdot20\right)\left(3\cdot4\cdot5\cdot\ldots\cdot21\right)}\)

\(C=\frac{22\cdot23}{20\cdot3}=\frac{506}{60}=\frac{253}{30}\)

20 tháng 10 2018

\(C=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{210}\right)\)

\(C=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{209}{210}\)

\(C=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{418}{420}\)

\(C=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...\frac{19.22}{20.21}\)

\(C=\frac{1.4.2.5.3.6.4.7...19.22}{2.3.3.4.4.5.5.6...20.21}\)

\(C=\frac{\left(1.2.3.4...19\right).\left(4.5.6.7...22\right)}{\left(2.3.4.5...20\right).\left(3.4.5.6...21\right)}\)

\(C=\frac{1.22}{20.3}=\frac{1.11}{10.3}=\frac{11}{30}\)

20 tháng 6 2018

\(a,A=\left(3\dfrac{5}{6}-1\dfrac{1}{3}\right)\left(3\dfrac{4}{15}-2\dfrac{3}{5}\right)\)
\(\Leftrightarrow A=\left(3+\dfrac{5}{6}-1+\dfrac{1}{3}\right)\left(3+\dfrac{4}{15}-2+\dfrac{3}{5}\right)\)
\(\Leftrightarrow A=\left[\left(3-1\right)+\left(\dfrac{5}{6}+\dfrac{1}{3}\right)\right]+\left[\left(3-2\right)+\left(\dfrac{4}{15}+\dfrac{3}{5}\right)\right]\)
\(\Leftrightarrow A=\left[2+\left(\dfrac{5}{6}+\dfrac{2}{6}\right)\right]+\left[1+\left(\dfrac{4}{15}+\dfrac{9}{15}\right)\right]\)
\(\Leftrightarrow A=\left(2+\dfrac{7}{6}\right)+\left(1+\dfrac{13}{15}\right)\)
\(\Leftrightarrow A=\left(2+1+\dfrac{1}{6}\right)+\left(1+\dfrac{13}{15}\right)\)
\(\Leftrightarrow A=3\dfrac{1}{6}+1\dfrac{13}{15}\)
Vậy...

20 tháng 6 2018

\(b,B=\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(\Leftrightarrow B=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.\left(2^3.3.5\right)}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(\Leftrightarrow B=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(\Leftrightarrow B=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(\Leftrightarrow B=\dfrac{\left(2^{10}.3^{10}\right)\left(1+5\right)}{\left(2^{11}.3^{11}\right)\left(2.3-1\right)}\)
\(\Leftrightarrow B=\dfrac{6}{\left(2.3\right).5}\)
\(\Leftrightarrow B=\dfrac{6}{6.5}\)
\(\Leftrightarrow B=\dfrac{1}{5}\)
Vậy....

3 tháng 3 2022

`Answer:`

\(C=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{210}\right)\)

\(=\left(\frac{3}{3}-\frac{1}{3}\right)\left(\frac{6}{6}-\frac{1}{6}\right)\left(\frac{10}{10}-\frac{1}{10}\right)\left(\frac{15}{15}-\frac{1}{15}\right)...\left(\frac{210}{210}-\frac{1}{210}\right)\)

\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{209}{210}\)

\(=\frac{2.2}{3.2}.\frac{5.2}{6.2}.\frac{9.2}{10.2}...\frac{209.2}{210.2}\)

\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{418}{420}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{19.22}{20.21}\)

\(=\frac{1.4.2.5.3.6...19.22}{2.3.3.4.4.5...20.21}\)

\(=\frac{\left(1.2.3...19\right)\left(4.5.6...22\right)}{\left(2.3.4...20\right)\left(3.4.5...21\right)}\)

\(=\frac{11}{30}\)

14 tháng 2 2015

C=2/3.5/6.9/10...209/210

C=4/6.10/12.18/20...418/420 là do nhân với 2

C=1.4/2.3.2.5/3.4.3.6/4.5...19.22/20.21

C=1.2.3....19/2.3.4...20.4.5.6...22/3.4.5...21

C=1/20.22/3

C=11/30

Dễ ấy mà hiểu chưa 

3 tháng 3 2020

C = \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)...\left(1-\frac{1}{210}\right)=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{209}{210}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{418}{420}\)

\(\frac{2.2}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{19.22}{20.21}=\frac{2.2\left(2.3.4...19\right)\left(5.6...22\right)}{\left(2.3.4..20\right)\left(3.4.5..21\right)}=\frac{4.22}{19.3.4}=\frac{22}{57}\)