Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Vì \(\frac{1}{33}>\frac{1}{34}>\frac{1}{35}>\frac{1}{36}\)
\(\Rightarrow M>\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\)\(\)
\(\Rightarrow M>\frac{4}{36}=\frac{1}{9}\)
Mà \(\frac{1}{9}>\frac{1}{10}\)
\(\Rightarrow\)\(M>\frac{1}{9}>\frac{1}{10}\)
Vậy : M > N
\(A=1+\frac{1}{2}+...+\frac{1}{16}\)
= \(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{12}\right)+\left(\frac{1}{13}+...+\frac{1}{16}\right)\)
> \(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+4\times\frac{1}{8}+4\times\frac{1}{12}+4\times\frac{1}{16}\)
=\(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
=\(1+2\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)
= \(1+2\times\frac{13}{12}\)
= \(1+\frac{13}{6}\)
= \(1+2+\frac{1}{6}\)
= \(3+\frac{1}{6}\)>\(3\)
=> \(A>3+\frac{1}{6}>3\)
=> \(A>3+\frac{1}{6}>B\)
=> \(A>B\)
\(\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{9}\right)\times\left(1-\frac{1}{16}\right)\times\left(1-\frac{1}{25}\right)\times\left(1-\frac{1}{36}\right)\)
\(=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times\frac{24}{25}\times\frac{36}{36}\)
\(=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times\frac{3.5}{4.4}\times\frac{4.6}{5.5}\times\frac{5.7}{6.6}\)
\(=\frac{1.2.3.4.5}{2.3.4.5.6}\times\frac{3.4.5.6.7}{2.3.4.5.6}\)
\(=\frac{1}{6}\times\frac{7}{2}\)
\(=\frac{7}{12}\)
(1-1/4)×(1-1/9)×(1-1/16)×(1-1/25)×(1-1/36)
=(4/4-1/4)×(9/9-1/9)×(16/16-1/16)×(25/25-1/25)×(36/36-1/36)
=3/4×8/9×15/16×24/25×35/36
=1×3×2×4×3×5×4×6×5×7/2×2×3×3×4×4×5×5×6×6
=(1×2×3×4×5)×(3×4×5×6×7)/(2×3×4×5×6)×(2×3×4×5×6)
=1/6×7/2
=7/12
Đề phải là \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}< \frac{1}{2}\) chứ ?
Ta có : \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}\)
\(=\frac{1}{4}\left(1+\frac{1}{4}+\frac{1}{9}+...+\frac{1}{2500}\right)\)
\(=\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Ta lại có : \(\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{4}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)
\(=\frac{1}{4}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=\frac{1}{4}\left(2-\frac{1}{50}\right)=\frac{1}{4}.\frac{99}{50}=\frac{99}{200}\)
Mà \(\frac{99}{200}< \frac{1}{2}\)\(\Rightarrow\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}< \frac{99}{200}< \frac{1}{2}\)
\(\Rightarrow\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}< \frac{1}{2}\) ( đpcm )
\(\text{Đặt BT là A }\Rightarrow A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(\text{Ta có:}\frac{1}{3^2}>\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\text{(để lại }\frac{1}{4}\text{ở đầu)}\)
\(\frac{1}{4^2}>\frac{1}{4.5}=\frac{1}{4}-\frac{1}{5}\)
.......
\(\frac{1}{100^2}>\frac{1}{100.101}\)
\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)
\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{101}\Rightarrow A=\frac{7}{12}-\frac{1}{101}=\frac{707-12}{1212}=\frac{695}{1212}>\frac{606}{1212}=\frac{1}{2}\)
\(\Rightarrow A>\frac{1}{2}\)
\(M=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
Ta thấy \(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2};\dfrac{1}{4^2}< \dfrac{1}{3\cdot4};...;\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\)
\(\Rightarrow M< \dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ =\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ =\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-1-\dfrac{1}{2}-...-\dfrac{1}{50}\\ =\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}\left(50.số\right)=\dfrac{50}{50}=1\)
Vậy \(M< 1\)
Mình chỉ so sánh với 1 được thôi à :((
Mình nghĩ cậu giải chưa đg đâu!:((