A = 124.( 1/1.1985 + 1/2.1986 + 1/3.1987 + … + 1/16.2000); B = 1/1.17 + 1/2.18 + 1/3.19 + … + 1/1984.2000. So sánh hai biểu thức A và B
A = 124.( 1/1.1985 + 1/2.1986 + 1/3.1987 + … + 1/16.2000); B = 1/1.17 + 1/2.18 + 1/3.19 + … + 1/1984.2000. So sánh hai biểu thức A và B
Đáp án:
Giải thích các bước giải:
\(A = 124 \cdot \left( {\frac{1}{{1.1985}} + \frac{1}{{2.1986}} + \frac{1}{{3.1987}} + … + \frac{1}{{16.2000}}} \right)\)
\( = \frac{{124}}{{1984}} \cdot \left( {\frac{{1984}}{{1.1985}} + \frac{{1984}}{{2.1986}} + \frac{{1984}}{{3.1987}} + … + \frac{{1984}}{{16.2000}}} \right)\)
\(= \frac{1}{{16}}\left( {1 – \frac{1}{{1985}} + \frac{1}{2} – \frac{1}{{1986}} + \frac{1}{3} – \frac{1}{{1987}} + … + \frac{1}{{16}} – \frac{1}{{2000}}} \right)\)
\( = \frac{1}{{16}} \cdot \left( {1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{{16}}} \right) \cdot \left( {\frac{1}{{1985}} + \frac{1}{{1986}} + \frac{1}{{1987}} + … + \frac{1}{{2000}}} \right)\)
\(B = \frac{1}{{1.17}} + \frac{1}{{2.18}} + … + \frac{1}{{1984.2000}}\)
\(= \frac{1}{{16}}\left( {\frac{{16}}{{1.17}} + \frac{{16}}{{2.18}} + … + \frac{{16}}{{1984.2000}}} \right)\)
\(= \frac{1}{{16}}\left( {1 – \frac{1}{{17}} + \frac{1}{2} – \frac{1}{{18}} + … + \frac{1}{{1984}} – \frac{1}{{2000}}} \right)\)
\( = \frac{1}{{16}}\left[ {\left( {1 + \frac{1}{2} + … + \frac{1}{{1984}}} \right) – \left( {\frac{1}{{17}} + \frac{1}{{18}} + … + \frac{1}{{2000}}} \right)} \right]\)
\( = \frac{1}{{16}}\left( {1 + \frac{1}{2} + … + \frac{1}{{16}}} \right) + \left( {\frac{1}{{17}} + \frac{1}{{18}} + … + \frac{1}{{1984}}} \right) – \left( {\frac{1}{{17}} + \frac{1}{{18}} + … + \frac{1}{{1984}}} \right) – \left( {\frac{1}{{1985}} + \frac{1}{{1986}} + … + \frac{1}{{2000}}} \right)\)
\(= \frac{1}{{16}}\left[ {\left( {1 + \frac{1}{2} + … + \frac{1}{{16}}} \right) – \left( {\frac{1}{{1985}} + \frac{1}{{1986}} + … + \frac{1}{{2000}}} \right)} \right]\)
Vậy \(A=B\).
Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}A = 124.\left( {\frac{1}{{1.1985}} + \frac{1}{{2.1986}} + \frac{1}{{3.1987}} + … + \frac{1}{{16.2000}}} \right)\\A = 124.\frac{1}{{1984}}.\left( {1 – \frac{1}{{1985}} + \frac{1}{2} – \frac{1}{{1986}} + \frac{1}{3} – \frac{1}{{1987}} + … + \frac{1}{{16}} – \frac{1}{{2000}}} \right)\\ = \frac{1}{{16}}.\left( {1 – \frac{1}{{1985}} + \frac{1}{2} – \frac{1}{{1986}} + \frac{1}{3} – \frac{1}{{1987}} + … + \frac{1}{{16}} – \frac{1}{{2000}}} \right)\\ = \frac{1}{{16}}.\left( {1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{{16}} – \left( {\frac{1}{{1985}} + \frac{1}{{1986}} + \frac{1}{{1987}} + … + \frac{1}{{2000}}} \right)} \right)\\B = \frac{1}{{1.17}} + \frac{1}{{2.18}} + \frac{1}{{3.19}} + …. + \frac{1}{{1984.2000}}\\ = \frac{1}{{16}}.\left( {1 – \frac{1}{{17}} + \frac{1}{2} – \frac{1}{{18}} + \frac{1}{3} – \frac{1}{{19}} + … + \frac{1}{{1984}} – \frac{1}{{2000}}} \right)\\ = \frac{1}{{16}}.\left( {1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{{1984}} – \frac{1}{{17}} – \frac{1}{{18}} – \frac{1}{{19}} – …. – \frac{1}{{2000}}} \right)\\ = \frac{1}{{16}}.\left( {1 + \frac{1}{2} + \frac{1}{3} + …. + \frac{1}{{1984}} – \left( {\frac{1}{{17}} + \frac{1}{{18}} + \frac{1}{{19}} + … + \frac{1}{{2000}}} \right)} \right)\\ = \frac{1}{{16}}\left[ {\left( {1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{{16}}} \right) + \left( {\frac{1}{{17}} + \frac{1}{{18}} + … + \frac{1}{{1984}}} \right) – \left( {\frac{1}{{17}} + \frac{1}{{18}} + … + \frac{1}{{1984}}} \right) – \left( {\frac{1}{{1985}} + \frac{1}{{1986}} + \frac{1}{{1987}} + … + \frac{1}{{2000}}} \right)} \right]\\ = \frac{1}{{16}}.\left[ {\left( {1 + \frac{1}{2} + … + \frac{1}{{16}}} \right) – \frac{1}{{1985}} + \frac{1}{{1986}} + … + \frac{1}{{2000}}} \right]\\ \Rightarrow A = B\end{array}\)