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\(F=\dfrac{49}{2.9}+\dfrac{49}{9.16}+............+\dfrac{49}{65.72}\)
\(\Leftrightarrow F=\dfrac{7^2}{2.9}+\dfrac{7^2}{9.16}+............+\dfrac{7^2}{65.72}\)
\(\Leftrightarrow F=7\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+.............+\dfrac{7}{65.72}\right)\)
\(\Leftrightarrow F=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...........+\dfrac{1}{65}-\dfrac{1}{75}\right)\)
\(\Leftrightarrow F=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\)
\(\Leftrightarrow F=7.\dfrac{35}{72}=\dfrac{245}{72}\)
\(G=\dfrac{3}{1.3}+\dfrac{3}{3.5}+...........+\dfrac{3}{47.49}\)
\(\Leftrightarrow G=\dfrac{3.2}{1.3.2}+\dfrac{3.2}{3.5.2}+........+\dfrac{3.2}{47.49}\)
\(\Leftrightarrow G=\dfrac{3}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+..........+\dfrac{2}{47.49}\right)\)
\(\Leftrightarrow G=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
\(\Leftrightarrow G=\dfrac{3}{2}\left(1-\dfrac{1}{49}\right)\)
\(\Leftrightarrow G=\dfrac{3}{2}.\dfrac{48}{49}=\dfrac{72}{49}\)
Ta có: \(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(P=\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+\left(1+\dfrac{3}{47}\right)+...+\left(1+\dfrac{48}{2}\right)+1\)
\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
\(P=50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(\Rightarrow\)\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}\)\(=\dfrac{1}{50}\)
\(Q=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{47}{3}+\dfrac{48}{2}+\dfrac{49}{1}\\ =\dfrac{1}{49}+1+\dfrac{2}{48}+1+\dfrac{3}{47}+1+...+\dfrac{47}{3}+1+\dfrac{48}{2}+1+1\\ =\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{3}+\dfrac{50}{2}+\dfrac{50}{50}\\ =50\cdot\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+...+\dfrac{1}{3}+\dfrac{1}{2}\right)\\ =50\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
\(\dfrac{P}{Q}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}}{50\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)
a) \(\dfrac{22}{55}=\dfrac{2}{5}\)
b) \(\dfrac{-63}{81}=\dfrac{-7}{9}\)
c) \(\dfrac{2.14}{7.8}=\dfrac{2.7.2}{7.2.2.2}=\dfrac{1}{2}\)
d) \(\dfrac{49+7.49}{49}=\dfrac{49.\left(7+1\right)}{49}=\dfrac{49.8}{49}=8\)
\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{49}{99}\)
\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{98}{99}\)
\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)
\(\dfrac{2x}{2x+1}=\dfrac{98}{99}\)
=> 2x=98
=> x=49
a. \(\dfrac{8,5-8,2}{16}=\dfrac{0,3}{16}=\dfrac{3}{160}\)
b. \(\dfrac{2\cdot14}{7\cdot8}=\dfrac{1\cdot2}{1\cdot4}=\dfrac{2}{4}=\dfrac{1}{2}\)
c. \(\dfrac{11\cdot4-11}{2-13}=\dfrac{11\left(4-1\right)}{-11}=\dfrac{1\cdot3}{-1}=-3\)
d. \(\dfrac{49+7\cdot49}{49}=\dfrac{49\cdot\left(1+7\right)}{49}=\dfrac{8}{1}=8\)
\(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(P=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+1\)
\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
\(P=50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)
\(B=\dfrac{49}{2\cdot9}+\dfrac{49}{9\cdot16}+\dfrac{49}{16\cdot23}+...+\dfrac{49}{65\cdot72}\)
\(B=\dfrac{49}{7}\cdot\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right)\)
\(B=7\cdot\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\)
\(B=7\cdot\left(\dfrac{36}{72}-\dfrac{1}{72}\right)\)
\(B=7\cdot\dfrac{35}{72}\)
\(B=\dfrac{\left(7\cdot35\right)}{72}\)
\(B=\dfrac{245}{72}\)
\(\dfrac{B}{7}=\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+\dfrac{7}{16\cdot23}+...+\dfrac{49}{65\cdot72}\\ \dfrac{B}{7}=\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\\ \dfrac{B}{7}=\dfrac{1}{2}-\dfrac{1}{72}\\ \dfrac{B}{7}=\dfrac{35}{72}\\ B=\dfrac{35}{72}\times7\\ B=\dfrac{245}{72} \)