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ko bt lm sao?!
Có tin t bảo cô m hỏi bài trên mạng không?
Mấy bài t hỏi là t đố con chính chủ xg con chính chủ nó đăng thôi
Ta có:
\(T=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.\dfrac{6}{5}.\dfrac{7}{6}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{3.4.5.6.7...99.100}{2.3.4.5.6...98.99}\)
\(=\dfrac{\left(3.4.5.6.7...99\right).100}{2.\left(3.4.5.6...98.99\right)}\)
\(=\dfrac{100}{2}=50\)
Vậy \(T=50\)
\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7} +.....................+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+....+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+..........+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+.......+\dfrac{1}{49.51}\right)}\)
\(=\dfrac{\dfrac{100}{1.99}+\dfrac{100}{3.97}+...........+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+...........+\dfrac{1}{49.51}\right)}\)
\(=\dfrac{100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+.............+\dfrac{1}{49.51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+..........+\dfrac{1}{49.51}\right)}\)
\(=\dfrac{100}{2}\)
\(=50\)
\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+.....+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+....+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+.....+\dfrac{1}{49.51}\right)}=\dfrac{\dfrac{100}{99}+\dfrac{100}{3.97}+....+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+....+\dfrac{1}{49.51}\right)}=\dfrac{100}{2}=50\)
\(T=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
Vậy T = 50
\(T=\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot\left(\dfrac{1}{4}+1\right)\cdot...\cdot\left(\dfrac{1}{98}+1\right)\cdot\left(\dfrac{1}{99}+1\right)\)
\(=\left(\dfrac{1}{2}+\dfrac{2}{2}\right)\cdot\left(\dfrac{1}{3}+\dfrac{3}{3}\right)\cdot\left(\dfrac{1}{4}+\dfrac{4}{4}\right)\cdot...\cdot\left(\dfrac{1}{98}+\dfrac{98}{98}\right)\cdot\left(\dfrac{1}{99}+\dfrac{99}{99}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{99}{98}\cdot\dfrac{100}{99}\)
\(=\dfrac{3\cdot4\cdot5\cdot...\cdot99\cdot100}{2\cdot3\cdot4\cdot...\cdot98\cdot99}\)
\(=\dfrac{100}{2}=50\).
Đặt :
\(A=\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^5}+.......................+\dfrac{1}{3^{99}}+\dfrac{1}{3^{99}}\)
\(\Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^5}+...................+\dfrac{1}{3^{98}}\)
\(\Rightarrow3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^5}+..............+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^3}+..............+\dfrac{1}{3^{98}}+\dfrac{1}{3^{99}}\right)\)\(\Rightarrow2A=1-\dfrac{1}{3^{99}}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}\)
\(\Rightarrow C=A+\dfrac{1}{8.3^{99}}=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}+\dfrac{1}{8.3^{99}}=\dfrac{1}{2}-\dfrac{3}{8.3^{99}}=\dfrac{1}{2}-\dfrac{1}{8.3^{98}}=\dfrac{4.3^{98}-1}{8.3^{98}}\)
Ta có: \(B=\left(3\dfrac{10}{99}+4\dfrac{11}{99}-5\dfrac{8}{299}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(B=\left(3\dfrac{10}{99}+4\dfrac{11}{99}-5\dfrac{8}{299}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)
\(B=\left(3\dfrac{10}{99}+4\dfrac{11}{99}-5\dfrac{8}{299}\right)\left(\dfrac{3}{6}+\dfrac{-2}{6}+\dfrac{-1}{6}\right)\)
\(B=\left(3\dfrac{10}{99}+4\dfrac{11}{99}-5\dfrac{8}{299}\right)\left(\dfrac{3+\left(-2\right)+\left(-1\right)}{6}\right)\)
\(B=\left(3\dfrac{10}{99}+4\dfrac{11}{99}-5\dfrac{8}{299}\right).0=0\)
Tick mk vs !
B = (3\(\dfrac{10}{99}\)+4\(\dfrac{11}{99}\)-5\(\dfrac{8}{299}\)).0
B = 0