\(=\left(\dfrac{4}{4}-\dfrac{1}{4}\right)\left(\dfrac{9}{9}-\dfrac{1}{9}\right)\left(\dfrac{16}{16}-\dfrac{1}{16}\right)...\left(\dfrac{10000}{10000}-\dfrac{1}{10000}\right)\)

\(=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}....\cdot\dfrac{9999}{10000}\)

\(=\dfrac{3.8.15.....9999}{4.9.16.....10000}=\dfrac{\left(1.3\right)\left(2.4\right)\left(3.5\right)....\left(99.101\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right).....\left(100.100\right)}\)

\(=\dfrac{\left(1.2.3...99\right)\left(3.4.5....101\right)}{\left(2.3.4...100\right)\left(2.3.4...101\right)}=\dfrac{101.1}{100.2}=\dfrac{101}{200}\)

\(E=\dfrac{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2002}-1\right)\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{9999}{10000}}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{100^2}\right)}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{100}{101}\cdot\dfrac{101}{102}\cdot...\cdot\dfrac{2002}{2003}}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{100}{2003}:\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\right)\)

\(=\dfrac{100}{2003}:\left(\dfrac{101}{2}\right)=\dfrac{100}{2003}\cdot\dfrac{2}{101}=\dfrac{200}{202303}\)

a)

\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)

\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)

\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)

\(=\dfrac{1}{100}.\dfrac{101}{2}\)

\(=\dfrac{101}{200}\)

ai bít câu b.c ko

ai giúp mìn vứi ❤

\(1-\dfrac{1}{n^2}=\dfrac{n^2-1}{n^2}=\dfrac{\left(n-1\right)\left(n+1\right)}{n^2}\)

\(\Rightarrow\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{n^2}\right)=\dfrac{1.3.2.4...\left(n-1\right)\left(n+1\right)}{2^2.3^2...n^2}\)

\(=\dfrac{1.2...\left(n-1\right)}{2.3...n}.\dfrac{3.4...\left(n+1\right)}{2.3...n}=\dfrac{1}{n}.\dfrac{n+1}{2}=\dfrac{n+1}{2n}\)

c)

Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)

\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)

d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)

\(=3-1+\dfrac{1}{4}:2\)

\(=3-1+\dfrac{1}{8}\)

\(=\dfrac{17}{8}\)