Tính giá trị biểu thức:a) \(A = \dfrac{9}{{2.5}} - \dfrac{{11}}{{3.5}} + \dfrac{{13}}{{3.7}} -

Câu hỏi số 577409:

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Tính giá trị biểu thức:

a) \(A = \dfrac{9}{{2.5}} - \dfrac{{11}}{{3.5}} + \dfrac{{13}}{{3.7}} - \dfrac{{15}}{{4.7}} + ... + \dfrac{{197}}{{49.99}} - \dfrac{{199}}{{50.99}}\)

b) \(B = 2022 + \dfrac{{2022}}{{1 + 2}} + \dfrac{{2022}}{{1 + 2 + 3}} + \dfrac{{2022}}{{1 + 2 + 3 + 4}} + ... + \dfrac{{2022}}{{1 + 2 + 3 + ... + 2021}}\)

c) \(C = \dfrac{{1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2021}}}}{{\dfrac{{2022}}{1} + \dfrac{{2023}}{2} + \dfrac{{2024}}{3} + ... + \dfrac{{4042}}{{2021}} - 2021}}\)

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Câu hỏi:577409

Phương pháp giải

+ \(S = \dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + ... + \dfrac{1}{{n\left( {n + 1} \right)}} = 1 - \dfrac{1}{{n + 1}} = \dfrac{n}{{n + 1}}\)

+ \(S = 1 + 2 + 3 + ... + n = \dfrac{{n\left( {n + 1} \right)}}{2}\)

Giải chi tiết

a) \(A = \dfrac{9}{{2.5}} - \dfrac{{11}}{{3.5}} + \dfrac{{13}}{{3.7}} - \dfrac{{15}}{{4.7}} + ... + \dfrac{{197}}{{49.99}} - \dfrac{{199}}{{50.99}}\)

\(\dfrac{A}{2} = \dfrac{1}{2}.\left( {\dfrac{9}{{2.5}} - \dfrac{{11}}{{3.5}} + \dfrac{{13}}{{3.7}} - \dfrac{{15}}{{4.7}} + ... + \dfrac{{197}}{{49.99}} - \dfrac{{199}}{{50.99}}} \right)\)

\(\begin{array}{l} = \dfrac{9}{{4.5}} - \dfrac{{11}}{{6.5}} + \dfrac{{13}}{{6.7}} - \dfrac{{15}}{{8.7}} + ... + \dfrac{{197}}{{98.99}} - \dfrac{{199}}{{100.99}}\\ = \dfrac{1}{4} + \dfrac{1}{5} - \left( {\dfrac{1}{6} + \dfrac{1}{5}} \right) + \dfrac{1}{6} + \dfrac{1}{7} - \left( {\dfrac{1}{8} + \dfrac{1}{7}} \right) + ... + \dfrac{1}{{98}} + \dfrac{1}{{99}} - \left( {\dfrac{1}{{100}} + \dfrac{1}{{99}}} \right)\\ = \dfrac{1}{4} + \dfrac{1}{5} - \dfrac{1}{6} - \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} - \dfrac{1}{8} - \dfrac{1}{7} + ... + \dfrac{1}{{98}} + \dfrac{1}{{99}} - \dfrac{1}{{100}} - \dfrac{1}{{99}}\\ = \dfrac{1}{4} - \dfrac{1}{{100}}\\ = \dfrac{6}{{25}}\end{array}\)

\( \Rightarrow A = \dfrac{6}{{25}}.2 = \dfrac{{12}}{{25}}\)

b) \(B = 2022 + \dfrac{{2022}}{{1 + 2}} + \dfrac{{2022}}{{1 + 2 + 3}} + \dfrac{{2022}}{{1 + 2 + 3 + 4}} + ... + \dfrac{{2022}}{{1 + 2 + 3 + ... + 2021}}\)

\(\begin{array}{l} = 2022 + 2022.\left( {\dfrac{1}{{1 + 2}} + \dfrac{1}{{1 + 2 + 3}} + \dfrac{1}{{1 + 2 + 3 + 4}} + ... + \dfrac{1}{{1 + 2 + 3 + ... + 2021}}} \right)\\ = 2022 + 2022.\left( {\dfrac{2}{{2\left( {1 + 2} \right)}} + \dfrac{2}{{2.\left( {1 + 2 + 3} \right)}} + \dfrac{2}{{2.\left( {1 + 2 + 3 + 4} \right)}} + ... + \dfrac{2}{{2.\left( {1 + 2 + 3 + ... + 2021} \right)}}} \right)\end{array}\)

\(\begin{array}{l} = 2022 + 2022.\left( {\dfrac{2}{6} + \dfrac{2}{{12}} + \dfrac{2}{{20}} + ... + \dfrac{2}{{2043231}}} \right)\\ = 2022 + 2022.2.\left( {\dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + \dfrac{1}{{4.5}} + ... + \dfrac{1}{{2021.2022}}} \right)\\ = 2022 + 4044.\left( {\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5} + ... + \dfrac{1}{{2021}} - \dfrac{1}{{2022}}} \right)\\ = 2022 + 4044.\left( {\dfrac{1}{2} - \dfrac{1}{{2022}}} \right)\\ = 2022 + 4044.\dfrac{{1010}}{{2022}}\\ = 2022 + 2020\\ = 4042\end{array}\)

c) \(C = \dfrac{{1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2021}}}}{{\dfrac{{2022}}{1} + \dfrac{{2023}}{2} + \dfrac{{2024}}{3} + ... + \dfrac{{4042}}{{2021}} - 2021}}\)

Xét mẫu \(\dfrac{{2022}}{1} + \dfrac{{2023}}{2} + \dfrac{{2024}}{3} + ... + \dfrac{{4042}}{{2021}} - 2021\)

\(\begin{array}{l} = \dfrac{{2021 + 1}}{1} + \dfrac{{2021 + 2}}{2} + \dfrac{{2021 + 3}}{3} + ... + \dfrac{{2021 + 2021}}{{2021}} - 2021\\ = 2021 + 1 + \dfrac{{2021}}{2} + 1 + \dfrac{{2021}}{2} + 1 + \dfrac{{2021}}{3} + 1 + ... + \dfrac{{2021}}{{2021}} + 1 - 2021\\ = 2021 + 1.2021 + \dfrac{{2021}}{2} + \dfrac{{2021}}{3} + \dfrac{{2021}}{4} + ... + \dfrac{{2021}}{{2021}} - 2021\\ = 2021 + \dfrac{{2021}}{2} + \dfrac{{2021}}{3} + \dfrac{{2021}}{4} + ... + \dfrac{{2021}}{{2021}}\\ = 2021.\left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + ... + \dfrac{1}{{2021}}} \right)\end{array}\)

\( \Rightarrow C = \dfrac{{1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2021}}}}{{2021\left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2021}}} \right)}} = \dfrac{1}{{2021}}\)

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