Question

[Nguyễn Đăng Nhân]

Biến đổi biểu thức sau:

\(\frac{5}{1\cdot6}=\frac{1}{?}+\frac{1}{?}\)

a) Từ đó, tính giá trị biểu thức:

\(A=\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+...+\frac{1}{2017\cdot2022}\)

b) Chứng minh \(B< A\)biết:

\(B=\frac{1}{6^2}+\frac{1}{11^2}+\frac{1}{16^2}+...+\frac{1}{2022^2}\)

Answer 1

[Phạm Nhật Đức]

a) \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+......+\frac{1}{2017.2022}\)

\(5A=5.\left(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+.....+\frac{1}{2017.2022}\right)\)

\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+......+\frac{5}{2017.2022}\)

\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+........+\frac{1}{2017}-\frac{1}{2022}\)

\(5A=1-\frac{1}{2022}\)

\(5A=\frac{2022}{2022}-\frac{1}{2022}\)

\(5A=\frac{2021}{2022}\)

\(A=\frac{2021}{2022}\div5\)

\(A=\frac{20201}{10110}\)

Answer 2

[(Tuấn) T̶r̶ư̶ở̶n̶g̶ 💎₮Ɇ₳₥ ₱ⱤØ🐬]

TL: 

\(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\) 

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HT