\(\frac{1}{255}+\frac{1}{323}+\frac{1}{399}+......+\frac{1}{9999}\)
Tính nhanh tổng sau : S = \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ \(\frac{1}{143}\)+ \(\frac{1}{195}\)+ \(\frac{1}{255}\)+ \(\frac{1}{323}\)
S = \(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{17x19}\)
2S = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\)\(\frac{1}{17}-\frac{1}{19}\)
2S = \(\frac{1}{3}-\frac{1}{19}\)
2S = \(\frac{16}{57}\)
S = \(\frac{16}{57}\times\frac{1}{2}\)
S = \(\frac{8}{57}\)
\(S=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}+\frac{1}{255}+\frac{1}{323}\)
\(S=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}+\frac{1}{15\cdot17}+\frac{1}{17\cdot19}\)
\(2S=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{15\cdot17}+\frac{2}{17\cdot19}\)
\(2S=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}\)
\(2S=\frac{1}{3}-\frac{1}{19}\)
\(2S=\frac{19}{57}-\frac{3}{57}\)
\(2S=\frac{16}{57}\)
\(S=\frac{16}{57}:2\)
\(S=\frac{16}{57}\cdot\frac{1}{2}\)
\(S=\frac{8}{57}\)
Ta có: S = 1/15 + 1/35 + 1/63 +...+1/323
=>S = 1/3.5 + 1/5.7 + 1/7.9 +...+ 1/17.19
=>S = 1/2(2/3.5 + 2/5.7 + 2/7.9 +...+ 2/17.19)
=>S = 1/2(1/3 - 1/5 + 1/5 - 1/7 + 1/7 -1/9 +...+ 1/17 - 1/19)
=>S = 1/2(1/3 - 1/19)
=>S = 1/2 . 16/57
=>S = 8/57.
Tính nhanh : \(\frac{10000}{10001}-\frac{9999}{10000}+\frac{1}{9999}-\frac{1}{10000}+...+\frac{3}{4}-\frac{2}{3}+\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{400\sqrt{399}+399\sqrt{400}}\)
Xét phân thức phụ sau:
Ta có: \(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+1\right)}}\cdot\left(\frac{1}{\sqrt{n}+\sqrt{n+1}}\right)\)
\(=\frac{1}{\sqrt{n\left(n+1\right)}}\cdot\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\frac{1}{\sqrt{n\left(n+1\right)}}\cdot\left(\sqrt{n+1}-\sqrt{n}\right)\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thay vào ta được:
\(BT=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\(BT=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)
Đặt biểu thức đã cho là A
Tổng quát ta có: Với \(a\inℕ^∗\)ta có:
\(\frac{1}{\left(a+1\right)\sqrt{a}+a.\sqrt{a+1}}=\frac{\left(a+1\right)-a}{\sqrt{a}.\sqrt{a+1}.\left(\sqrt{a}+\sqrt{a+1}\right)}\)
\(=\frac{\left(\sqrt{a+1}-\sqrt{a}\right)\left(\sqrt{a+1}+\sqrt{a}\right)}{\sqrt{a}.\sqrt{a+1}.\left(\sqrt{a}+\sqrt{a+1}\right)}=\frac{\sqrt{a+1}-\sqrt{a}}{\sqrt{a}.\sqrt{a+1}}\)
\(=\frac{\sqrt{a+1}}{\sqrt{a}.\sqrt{a+1}}-\frac{\sqrt{a}}{\sqrt{a}.\sqrt{a+1}}=\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a+1}}\)
Áp dụng kết quả trên ta có:
Với \(n=1\)\(\Rightarrow\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)
Với \(n=2\)\(\Rightarrow\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
Với \(n=3\)\(\Rightarrow\frac{1}{4\sqrt{3}+3\sqrt{4}}=\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\)
.....................
Với \(n=399\)\(\Rightarrow\frac{1}{400\sqrt{399}+399\sqrt{400}}=\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\(\Rightarrow A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+......+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)
\(CMR:\frac{1}{201}+\frac{1}{202}+....+\frac{1}{399}+\frac{1}{400}>\frac{1}{2}\)
Đặt \(S=\frac{1}{201}+\frac{1}{202}+...+\frac{1}{399}+\frac{1}{400}\)
Ta thấy :
\(\frac{1}{201}>\frac{1}{400}\)
\(\frac{1}{202}>\frac{1}{400}\)
...
\(\frac{1}{399}>\frac{1}{400}\)
\(\Rightarrow S>\frac{1}{400}+\frac{1}{400}+\frac{1}{400}+...+\frac{1}{400}\)
có 200 dãy \(\Rightarrow S>\frac{200}{400}=\frac{1}{2}\)
Vậy : \(S>\frac{1}{2}\)
Chứng tỏ rằng:
\(\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{399}+\frac{1}{400}>\frac{1}{2}\)
Vì \(\frac{1}{201}>\frac{1}{400}\)
\(\frac{1}{202}>\frac{1}{400}\)
\(\frac{1}{203}>\frac{1}{400}\)
.................
\(\frac{1}{399}>\frac{1}{400}\)
⇒ \(\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{399}>\frac{1}{400}+\frac{1}{400}+\frac{1}{400}+...+\frac{1}{400}\)(199 số hạng \(\frac{1}{400}\))
⇒ \(\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{399}+\frac{1}{400}>\frac{1}{400}+\frac{1}{400}+\frac{1}{400}+...+\frac{1}{400}\)(200 số hạng \(\frac{1}{400}\)) = 200.\(\frac{1}{400}\)=\(\frac{1}{2}\)
⇒ A > \(\frac{1}{2}\)
Vậy A > \(\frac{1}{2}\) (ĐPCM)
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}.\frac{98}{303}\)
\(A=\frac{49}{303}\)
A= \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
2A=\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
2A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
2A=\(\frac{1}{3}-\frac{1}{101}\)
2A=\(\frac{98}{303}\)
A=\(\frac{98}{303}.\frac{1}{2}\)
A=\(\frac{49}{303}\)
Chúc bạn học tốt!
tìm x : \(\frac{1}{x}-\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{99}\right)\)
\(\frac{1}{x}-\frac{1}{999}=\frac{1}{2}.\frac{98}{99}\)
\(\frac{1}{x}-\frac{1}{9999}=\frac{49}{99}\)
\(\frac{1}{x}=\frac{49}{99}+\frac{1}{9999}\)
\(\frac{1}{x}=\frac{50}{101}\)
\(x=1:\frac{50}{101}\)
\(x=\frac{101}{50}\)
Vậy \(x=\frac{101}{50}\)
\(\frac{1}{3}+\frac{1}{5}+\frac{1}{35}+...+\frac{1}{9999}\)
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}=?\)