Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Đặt :
\(A=\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+.................+\dfrac{1}{100!}\)
Ta thấy :
\(\dfrac{1}{2!}=\dfrac{1}{1.2}\)
\(\dfrac{1}{3!}=\dfrac{1}{1.2.3}\)
\(\dfrac{1}{4!}=\dfrac{1}{1.2.3.4}< \dfrac{1}{3.4}\)
.....................................
\(\dfrac{1}{100!}=\dfrac{1}{1.2.3..........100}< \dfrac{1}{99.100}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...........+\dfrac{1}{99.100}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A< 1-\dfrac{1}{100}\)
\(A< \dfrac{99}{100}< 1\)
\(\Rightarrow A< 1\rightarrowđpcm\)
b) Đặt :
\(B=\dfrac{9}{10!}+\dfrac{9}{11!}+\dfrac{9}{12!}+.............+\dfrac{9}{1000!}\)
Ta thấy :
\(\dfrac{9}{10!}=\dfrac{10-1}{10!}=\dfrac{1}{9!}-\dfrac{1}{10!}\)
\(\dfrac{9}{11!}< \dfrac{11-1}{11!}=\dfrac{1}{10!}-\dfrac{1}{11!}\)
...................................................
\(\dfrac{9}{1000!}< \dfrac{1000-1}{1000!}=\dfrac{1}{999!}-\dfrac{1}{1000!}\)
\(\Rightarrow B< \dfrac{1}{9!}-\dfrac{1}{10!}+\dfrac{1}{10!}-\dfrac{1}{11!}+............+\dfrac{1}{999!}-\dfrac{1}{1000!}\)
\(B< \dfrac{1}{9!}-\dfrac{1}{1000!}\)
\(\Rightarrow B< \dfrac{1}{9!}\rightarrowđpcm\)
~ Chúc bn học tốt ~
Ta có:
\(\dfrac{9}{n!}\)< \(\dfrac{n-1}{n!}\) = \(\dfrac{1}{(n-1)!} - \dfrac{1}{n!}\) với n > 10 (n thuộc Z)
\(\Rightarrow\) \(\dfrac{9}{10!} + \dfrac{9}{11!} + \dfrac{9}{12!} + ... +\dfrac{9}{1000!} \)
= \(\dfrac{1}{9!} - \dfrac{1}{10!} + \dfrac{9}{11!} + \dfrac{9}{12!} + ... +\dfrac{9}{1000!}\)
\(\Rightarrow\) \(\dfrac{1}{9!} - \dfrac{1}{10!} + \dfrac{1}{10!} - \dfrac{1}{11!} + \dfrac{1}{11!} - \dfrac{1}{12!} + ....\)
= \(\dfrac{1}{9!} - \dfrac{1}{1000!}\)
\(\Rightarrow \) \(\dfrac{9}{10!} + \dfrac{9}{11!} + ...+ \dfrac{9}{1000!} < \dfrac{1}{9!}\)
Chúc bn hc tốt.
N = \(\dfrac{1}{10^2}+\dfrac{1}{11^2}+\dfrac{1}{12^2}+...+\dfrac{1}{n^2}\)
= \(\dfrac{1}{10.10}+\dfrac{1}{11.11}+\dfrac{1}{12.12}+...+\dfrac{1}{n.n}\)
=> N < \(\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}+...+\dfrac{1}{\left(n-1\right).n}\)
=> N < \(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(=>N< \dfrac{1}{9}-\dfrac{1}{n}\)
=> N < \(\dfrac{1}{9}\)
Vậy N < \(\dfrac{1}{9}\)
a)Ta có:\(\dfrac{1}{b}-\dfrac{1}{b+1}=\dfrac{b+1-b}{b\left(b+1\right)}=\dfrac{1}{b^2+b}< \dfrac{1}{b^2}\)(do b>1)
\(\dfrac{1}{b-1}-\dfrac{1}{b}=\dfrac{b-b+1}{\left(b-1\right)b}=\dfrac{1}{b^2-b}>\dfrac{1}{b^2}\)(do b>1)
b)Áp dụng từ câu a
=>\(\dfrac{1}{2}-\dfrac{1}{3}< \dfrac{1}{2^2}< \dfrac{1}{1}-\dfrac{1}{2}\)
\(\dfrac{1}{3}-\dfrac{1}{4}< \dfrac{1}{3^2}< \dfrac{1}{2}-\dfrac{1}{3}\)
.........................
\(\dfrac{1}{9}-\dfrac{1}{10}< \dfrac{1}{9^2}< \dfrac{1}{8}-\dfrac{1}{9}\)
=>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}< S< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
=>\(\dfrac{1}{2}-\dfrac{1}{10}< S< 1-\dfrac{1}{9}\)
=>\(\dfrac{2}{5}< S< \dfrac{8}{9}\)(đpcm)
Câu a :
Chưa nghĩ ra! Sorry nhé!!
Câu b :
Câu hỏi của Trần Thùy Linh - Toán lớp 6 | Học trực tuyến
Câu c :
Câu hỏi của Trần Thùy Linh - Toán lớp 6 | Học trực tuyến
Vào link đó mà xem, t ngại chép lại
\(A=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+....+\dfrac{\left(3^{99}+1\right)}{3^{99}}\)
\(A=\dfrac{4}{3}+\dfrac{10}{3^2}+\dfrac{28}{3^3}+...+\dfrac{\left(3^{99}+1\right)}{3^{99}}\)
\(A=\left(1+\dfrac{1}{3}\right)+\left(1+\dfrac{1}{3^2}\right)+\left(1+\dfrac{1}{3^3}\right)+...+\left(1+\dfrac{1}{3^{99}}\right)\)
\(A=\left(1+1+....+1\right)+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(A=99+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)
Gọi \(\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)là T
\(T=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)
\(3T=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\)
\(3T-T=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)
\(2T=1-\dfrac{1}{3^{99}}\)
\(T=\left(1-\dfrac{1}{3^{99}}\right):2\)
\(T=\dfrac{1}{2}-\dfrac{1}{3^{99}\cdot2}\)
\(=>A=99+T=99+\dfrac{1}{2}-\dfrac{1}{3^{99}\cdot2}=99,5-\dfrac{1}{3^{99}\cdot2}< 100\)
Vậy A < 100
Ta có:
1/2^2 > 1/2.3
1/3^2 > 1/3.4
...
1/10^2 > 1/10.11
-> Cộng dọc theo vế ta có:
1/2^2+1/3^2+...+1/10^2 > 1/2.3+1/3.4+...+1/10.11
= 1/2-1/3+1/3-1/4+...+1/10-1/11
= 1/2 - 1/11 = 9/22 (đpcm)
b)B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
B<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
B<\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
B<\(1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+...+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)-\dfrac{1}{9}\)
B<1-\(\dfrac{1}{9}\)
B<\(\dfrac{8}{9}\)(1)
ta có:
B>\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
B>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{10}\)
B>\(\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)...+\left(\dfrac{1}{9}+\dfrac{1}{9}\right)-\dfrac{1}{10}\)
B>\(\dfrac{1}{2}-\dfrac{1}{10}\)
B>\(\dfrac{2}{5}\)
a, Ta có :
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{1\cdot2\cdot3\cdot4}+...+\dfrac{1}{1\cdot2\cdot3\cdot...\cdot100}\\ < \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}< 1\\ \Rightarrow M< 1\\ \RightarrowĐpcm\)