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Có:
\(A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{81}+\dfrac{1}{100}\)
\(A=\dfrac{1}{4}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}+\dfrac{1}{10^2}\)
Mà: \(\dfrac{1}{3^2}>\dfrac{1}{3.4}\)
\(\dfrac{1}{4^2}>\dfrac{1}{4.5}\)
...
\(\dfrac{1}{9^2}>\dfrac{1}{9.10}\)
\(\dfrac{1}{10^2}>\dfrac{1}{10.11}\)
\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)
\(A>\dfrac{1}{4}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)
\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{3}-0-0-...-0-\dfrac{1}{11}\)
\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{11}\)
\(\Rightarrow A>\dfrac{65}{132}\)
Chúc bạn học tốt!
Đặt \(A=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+...+\dfrac{1}{196}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{13^2}\)
Đặt \(B=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{12\cdot13}\)
Ta có:
\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{13^2}\)\(<\)\(B=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{12\cdot13}\left(1\right)\)
Mà \(B=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{12\cdot13}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{12}-\dfrac{1}{13}\)
\(=\dfrac{1}{2}-\dfrac{1}{13}< \dfrac{1}{2}\left(2\right)\). Từ \((1)\) và \((2)\) ta có:
\(A< B< \dfrac{1}{2}\Rightarrow A< \dfrac{1}{2}\) (Điều phải chứng minh)
a) Giải
Đặt \(M=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\)
\(\Rightarrow A< A.M\)
hay \(A< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\right)\)
\(\Rightarrow A< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}.\dfrac{6}{7}...\dfrac{98}{99}.\dfrac{99}{100}\)
\(\Leftrightarrow A< \dfrac{1.2.3.4.5.6...98.99}{2.3.4.5.6.7...99.100}\)
\(\Rightarrow A< \dfrac{1}{100}< \dfrac{1}{10}\)
Vậy \(A< \dfrac{1}{10}\)
bài 2
| a;đặt biểu thức là S | |
| S < 1/1.2 + 1/2.3 + .......1/(n-1)n | |
| = 1- 1/2 +1 /2 -1/3+........ + 1/n-1 - 1/n | |
|
= 1 -1/n <1 |
|
| vậy S < 1 | |
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)...\left(\dfrac{1}{100}-1\right)\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{8}{9}\right).\left(-\dfrac{15}{16}\right)...\left(-\dfrac{99}{100}\right)\)
\(A=\dfrac{\left(-1\right).3}{2^2}.\dfrac{\left(-2\right).4}{3^2}.\dfrac{\left(-3\right).5}{4^2}....\dfrac{\left(-9\right).11}{10^2}\)
\(A=\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)....\left(-9\right)}{2.3.4....10}.\dfrac{3.4.5....11}{2.3.4....10}\)
\(A=\dfrac{-1}{10}.\dfrac{11}{2}=-\dfrac{11}{20}\)
Câu B tương tự nha bạn!!!
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)...\left(\dfrac{1}{100}-1\right)\)
\(A=\dfrac{-3}{4}.\dfrac{-8}{9}.\dfrac{-15}{16}......\dfrac{-99}{100}\)
\(A=\dfrac{-1.3}{2.2}.\dfrac{-2.4}{3.3}.\dfrac{-3.5}{4.4}.....\dfrac{-9.11}{10.10}\)
\(A=\dfrac{-1.3.-2.4.-3.5.....-9.11}{2.2.3.3.4.4.....10.10}\)
\(A=\dfrac{-1.-2.-3......-9}{2.3.4......10}.\dfrac{3.4.5....11}{2.3.4...10}\)
\(A=\dfrac{-1}{10}.\dfrac{11}{2}=\dfrac{-11}{20}\)
\(B=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right).....\left(\dfrac{1}{10}-1\right)\)
\(B=\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}........\dfrac{-9}{10}\)
\(B=\dfrac{-1.-2.-3......-9}{2.3.4......10}\)
\(B=\dfrac{-1}{10}\)
Cho A = 1/2 .3/4.5/6.....199/200.Chứng tỏ rằng B mũ 2 <1/201.Bạn có làm dược ko ?