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Ta có:
\(A=\dfrac{1}{1.1981}+\dfrac{1}{2.1982}+...+\dfrac{1}{n\left(1980+n\right)}+...+\dfrac{1}{25.2005}\)
\(=\dfrac{1}{1980}\left(\dfrac{1981-1}{1.1981}+\dfrac{1982-2}{2.1982}+...+\dfrac{1980+n-n}{n\left(1980+n\right)}+...+\dfrac{2005-25}{25.2005}\right)\)
\(=\dfrac{1}{1980}\left(1-\dfrac{1}{1981}+\dfrac{1}{2}-\dfrac{1}{1982}+...+\dfrac{1}{n}-\dfrac{1}{1980+n}+...+\dfrac{1}{25}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{1980}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)
Lại có:
\(B=\dfrac{1}{1.26}+\dfrac{1}{2.27}+...+\dfrac{1}{m\left(m+25\right)}+...+\dfrac{1}{1980.2005}\)
\(=\dfrac{1}{25}\left(\dfrac{26-1}{1.26}+\dfrac{27-2}{2.27}+...+\dfrac{25+m-m}{m\left(25+m\right)}+...+\dfrac{2005-1980}{1980.2005}\right)\)
\(=\dfrac{1}{25}\left(\dfrac{1}{1}-\dfrac{1}{26}+\dfrac{1}{2}-\dfrac{1}{27}+...+\dfrac{1}{m}-\dfrac{1}{25+m}+...+\dfrac{1}{1980}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{25}\left[\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{1980}\right)-\left(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{2005}\right)\right]\)
\(=\dfrac{1}{25}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{1980}}{\dfrac{1}{25}}=\dfrac{5}{396}\)
Vậy tỉ số của \(A\) và \(B\) là \(\dfrac{5}{396}\)
25B=\(\frac{25}{1.26}+\frac{25}{2.27}+.......+\frac{25}{1980.2005}\)
\(=1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+..........+\frac{1}{1980}-\frac{1}{2005}\)
\(=1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{25}-\frac{1}{1981}-\frac{1}{1982}-....-\frac{1}{2005}\)
1980A=\(\frac{1980}{1.1981}+\frac{1980}{2.1982}+.........+\frac{1980}{25.2005}\)
\(=1-\frac{1}{1981}+\frac{1}{2}-\frac{1}{1982}+......+\frac{1}{25}-\frac{1}{2005}\)
\(=1+\frac{1}{2}+...+\frac{1}{25}-\frac{1}{1981}-\frac{1}{1982}-...-\frac{1}{2005}\)
\(\Rightarrow25B=1980A\)\(\Rightarrow\frac{A}{B}=\frac{25}{1980}\)
a) \(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right):\left(-3\right)\)
\(A=\left(-0,75-0,25\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right)\cdot\dfrac{-1}{3}\)
\(A=\left(-1\right):\left(-5\right)+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{1}{5}+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{119}{720}\)
b) \(B=\left(\dfrac{6}{25}-1,24\right):\dfrac{3}{7}:\left[\left(3\dfrac{1}{2}-3\dfrac{2}{3}\right):\dfrac{1}{14}\right]\)
\(B=\left(0,24-1,24\right):\dfrac{3}{7}:\left[\left(\dfrac{7}{2}-\dfrac{11}{3}\right):\dfrac{1}{14}\right]\)
\(B=-1:\dfrac{3}{7}:\left(-\dfrac{1}{6}:\dfrac{1}{14}\right)\)
\(B=-\dfrac{7}{3}:-\dfrac{7}{3}\)
\(B=1\)
a, A = (-0,75 - \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{48}\) - (- \(\dfrac{1}{6}\)) : (-3)
A = -(0,75 + 0,25): (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = -1 : (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{1}{5}\) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{53}{240}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{119}{720}\)
b, B = (\(\dfrac{6}{25}\) - 1,24): \(\dfrac{3}{7}\): [(3\(\dfrac{1}{2}\) - 3\(\dfrac{2}{3}\)): \(\dfrac{1}{14}\)]
B = (0,24 - 1,24): \(\dfrac{3}{7}\):[(\(\dfrac{7}{2}\)-\(\dfrac{11}{3}\)): \(\dfrac{1}{14}\)]
B = -1: \(\dfrac{3}{7}\):[ (-\(\dfrac{1}{6}\) : \(\dfrac{1}{14}\))]
B = -1: \(\dfrac{3}{7}\): (- \(\dfrac{7}{3}\))
B = 1 \(\times\) \(\dfrac{7}{3}\) \(\times\) \(\dfrac{3}{7}\)
B = 1
\(a.=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}=\dfrac{1+3-5}{2}-\dfrac{2+5-7}{3}=\dfrac{-1}{2}\)
\(b.\left(\dfrac{3}{4}-1\dfrac{1}{6}\right)^2:\sqrt{\dfrac{25}{144}}=\left(-\dfrac{5}{12}\right)^2:\dfrac{5}{12}=\dfrac{5}{12}\)
Lời giải:
Ta có \(A=\frac{1}{1.1981}+\frac{1}{2.1982}+...+\frac{1}{25.2005}\)
\(\Rightarrow 1980A=\frac{1980}{1.1981}+\frac{1980}{2.1982}+...+\frac{1980}{25.2005}\)
\(\Leftrightarrow 1980A=\frac{1981-1}{1.1981}+\frac{1982-2}{2.1982}+....+\frac{2005-25}{25.2005}\)
\(\Leftrightarrow 1980A=1-\frac{1}{1981}+\frac{1}{2}-\frac{1}{1982}+...+\frac{1}{25}-\frac{1}{2005}\)
\(1980A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)-\left(\frac{1}{1981}+\frac{1}{1982}+..+\frac{1}{2005}\right)\) (1)
Lại có:
\(25B=\frac{25}{1.26}+\frac{25}{2.27}+...+\frac{25}{1980.2005}\)
\(\Leftrightarrow 25B=\frac{26-1}{1.26}+\frac{27-2}{2.27}+...+\frac{2005-1980}{1980.2005}\)
\(\Leftrightarrow 25B=1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+...+\frac{1}{1980}-\frac{1}{2005}\)
\(25B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1980}\right)-\left(\frac{1}{26}+\frac{1}{27}+....+\frac{1}{2005}\right)\)
\(25B=\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{1981}+\frac{1}{1982}+...+\frac{1}{2005}\right)\) (2)
Từ \((1); (2)\Rightarrow 1980A=25B\Rightarrow \frac{A}{B}=\frac{25}{1980}=\frac{5}{396}\)