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\(Q=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)
\(Q=\left(\frac{1}{2}\right).\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)...\left(\frac{99}{100}\right)\)
\(Q=\frac{1}{100}\)
\(P=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(P=\left(\frac{1.3}{1.3}+\frac{1}{1.3}\right)\left(\frac{2.4}{2.4}+\frac{1}{2.4}\right)\left(\frac{3.5}{3.5}+\frac{1}{3.5}\right)...\left(\frac{99.101}{99.101}+\frac{1}{99.101}\right)\)
\(P=\left(\frac{4}{1.3}\right)\left(\frac{9}{2.4}\right)\left(\frac{16}{3.5}\right)...\left(\frac{10000}{99.101}\right)\)
\(P=\left(\frac{2^2}{1.3}\right)\left(\frac{3^2}{2.4}\right)\left(\frac{4^2}{3.5}\right)...\left(\frac{100^2}{99.101}\right)\)
Bạn tự tách ra rồi bạn sẽ ra kết quả như ở dưới
\(P=\frac{201}{100}\)
\(\frac{1}{1.3}.\frac{1}{3.5}.\frac{1}{5.6}.....\frac{1}{99.100}\)
\(=\frac{1}{1}.\left(\frac{1}{3.3}\right).\left(\frac{1}{5.5}\right).\left(\frac{1}{6.6}\right).....\left(\frac{1}{99.99}\right).\frac{1}{100}\)
\(=\frac{1}{1}.1.1.1.....1.\frac{1}{100}\)
\(=\frac{1}{100}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{98}-\frac{1}{100}\right)\)\(=\frac{1}{2}.\left(1-\frac{1}{100}\right)=\frac{1}{2}.\frac{99}{100}=\frac{99}{200}\)
\(C=\left[1+\frac{1}{1\cdot3}\right]\left[1+\frac{1}{2\cdot4}\right]...\left[1+\frac{1}{2014\cdot2016}\right]\)
\(=\frac{4}{3}\cdot\frac{9}{8}\cdot\frac{16}{15}\cdot...\cdot\frac{4060225}{4060224}\)
\(=\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\frac{4\cdot4}{3\cdot5}\cdot...\cdot\frac{2015\cdot2015}{2014\cdot2016}\)
\(=\frac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot2015\cdot2015}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot2014\cdot2016}\)
Để ý kĩ thì các thừa số dưới mẫu so với trên tử giống nhau chỉ khác 2016 nên C bằng:
C = 2*2*3*3*4*4*...*2015*2015/1*2*3*3*4*4*5*5*...*2015*2015*2016 = 1/2016
Ta có : (a-1)(a+1)=a2+a-a-1=a2-1
\(\Rightarrow\)(a-1)(a+1)+1=a2
Từ đó ta có :
\(C=\frac{2^2}{1.3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot...\cdot\frac{2015^2}{2014\cdot2016}\)
\(\Rightarrow\)\(C=\left(\frac{2\cdot3\cdot4\cdot...\cdot2015}{1\cdot2\cdot3\cdot...\cdot2014}\right)\cdot\left(\frac{2\cdot3\cdot4\cdot...2015}{3\cdot4\cdot5\cdot...\cdot2016}\right)\)
\(\Rightarrow\)\(C=\frac{2015}{1}\cdot\frac{1}{2016}\)
\(\Rightarrow\)\(C=\frac{2015}{2016}\)
\(A=\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.100}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{3}-\frac{1}{100}=\frac{97}{300}\)
Ta có:
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...\frac{9}{98.99}+\frac{9}{99.100}\)
\(=9.\frac{1}{1.2}+9.\frac{1}{2.3}+9.\frac{1}{3.4}+...+9.\frac{1}{98.99}+9.\frac{1}{99.100}\)
\(=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=9.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=9.\left(1-\frac{1}{100}\right)\)
\(=9.\frac{99}{100}\)
\(=\frac{9.99}{100}\)
\(=\frac{891}{100}\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=9.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(M=\frac{\left(2.4....98.100\right).\left(51.52....99.100\right)}{\left(2.4....98.100\right).\left(1.3....97.99\right).2^{50}}\)
\(M=\frac{\left(2.1\right).\left(2.3\right)....\left(2.49\right).\left(2.50\right).51.52....99.100}{1.2......99.100.2^{50}}\)
\(M=\frac{2^{50}.\left(1.2....99.100\right)}{\left(1.2....99.100\right).2^{50}}\)\(=1\)
Vậy M =1
Chúc bạn học tốt ( -_- )
\(M=\frac{51.52...99.100}{1.3...97.99.2^{50}}\)
\(=\frac{\left(1.2.3...50\right).51.52...99.100}{1.3...97.99.2^{50}.\left(1.2.3...50\right)}\)
\(=\frac{1.2.3...99.100}{1.3...97.99.\left(2.4.6...100\right)}\)
\(=\frac{1.2.3...99.100}{1.2.3...99.100}\)