1 nha bạn

k mik nhé

\(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right):\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)=1\)

(1/6 + 1/10 + 1/15) : (1/6 +1/10 +1/15)=1 

Chúc bạn học tốt!

\(B=\left(\frac{1}{3}-1\right).\left(\frac{1}{6}-1\right).\left(\frac{1}{10}-1\right).......\left(\frac{1}{1225}-1\right)\left(\frac{1}{1275}-1\right)\)

\(B=\frac{-2}{3}.\frac{-5}{6}.\frac{-9}{10}......\frac{-1224}{1225}.\frac{-1274}{1275}\)

\(B=\frac{-4}{6}.\frac{-10}{12}.\frac{-18}{20}......\frac{-2448}{2450}.\frac{-2548}{2550}\)

\(B=\frac{-4}{2.3}.\frac{-10}{3.4}.\frac{-18}{4.5}.....\frac{-2448}{49.50}.\frac{-2548}{50.51}\)

\(\Rightarrow\)B có : ( 50 - 2 ) : 1 + 1 = 49 ( số hạng )

\(\Rightarrow B=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}........\frac{2448}{49.50}.\frac{2548}{50.51}.\left(-1\right)\)

     \(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.........\frac{48.51}{49.50}.\frac{49.52}{50.51}.\left(-1\right)\)

     \(B=\frac{\left(1.2.3...48.49\right).\left(4.5.6......51.52\right)}{\left(2.3.4......49.50\right).\left(3.4.5.....50.51\right)}.\left(-1\right)\)

     \(B=\frac{52}{50.3}.\left(-1\right)\)

      \(B=\frac{26}{75}.\left(-1\right)\)

Vậy \(B=\frac{-26}{75}\)

sửa đề :

\(C=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{15}\right)+...+\left(1-\frac{1}{105}\right)\)

\(C=\left(1+1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{105}\right)\)

Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{105}\)

\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{210}\)

\(A=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{210}\right)\)

\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{14.15}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{14}-\frac{1}{15}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{15}\right)\)

\(A=2.\frac{13}{30}\)

\(A=\frac{13}{15}\)

Thay A vào ta được :

B = \(\left(1+1+1+1+...+1\right)-\frac{13}{15}\)

B = \(14-\frac{13}{15}\)( có 14 số 1 )

B = \(\frac{197}{15}\)

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

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

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

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

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

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

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

\(=\frac{-2}{3}\cdot\frac{-5}{6}\cdot\frac{-9}{10}\cdot\cdot\cdot\cdot\frac{-35}{36}\)

\(=\frac{-4}{6}\cdot\frac{-10}{12}\cdot\frac{-18}{20}\cdot\cdot\cdot\cdot\frac{-70}{72}\)

\(=\frac{-1.4}{2.3}\cdot\frac{-2.5}{3.4}\cdot\frac{-3.6}{4.5}\cdot\cdot\cdot\cdot\frac{-7.10}{8.9}\)

\(=\frac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-7\right)}{2.3.4....8}\cdot\frac{4.5.6....10}{3.4.5....9}\)

\(=\frac{\left(-1\right).2.3...7}{2.3.4....8}\cdot\frac{10}{3}\)

\(=\frac{-1}{8}\cdot\frac{10}{3}=\frac{-5}{12}\)

\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right)\)

\(\Rightarrow B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{779}{780}\)

\(\Rightarrow B=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{1558}{1560}\)

\(\Rightarrow B=\frac{1.4}{2.3}.\frac{2.5}{3.4}\frac{3.6}{4.5}...\frac{38.41}{39.40}\)

\(\Rightarrow B=\frac{\left(1.2.3...38\right)\left(4.5.6...41\right)}{\left(2.3.4...39\right)\left(3.4.5...40\right)}\)

\(\Rightarrow B=\frac{1.41}{39.3}=\frac{41}{117}\)

Vậy B=\(\frac{41}{117}\)

Ai thấy đúng thì k nha

B= \(\frac{41}{117}\)