\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{99}{100}\)
K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

3 tháng 8 2023

\(\Leftrightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{99}{100}\)

\(\Leftrightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{99}{100}\)

\(\Leftrightarrow1-\dfrac{1}{n+1}=\dfrac{99}{100}\)

\(\Leftrightarrow\dfrac{1}{n+1}=\dfrac{1}{100}\)

=> n+1=100

n=99

3 tháng 8 2023

1/2 + 1/6 + 1/12 + ... + 1/(n(n+1)) = 99/100`

`=> 1/(1.2) + 1/(2.3) + 1/(3.4) + ... + 1/(n.(n+1)) = 99/100`

`=> 1/1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 + ... + 1/n - 1/(n+1) = 99/100`

`=> 1 - 1/(n+1) = 99/100`

`=>      1/(n + 1) = 1 - 99/100`

`=>    1/(n + 1 ) = 1/100`

`=>       n  +1   = 100`

`=>      n          =  99`

19 tháng 6 2017

\(\dfrac{100+\dfrac{99}{2}+\dfrac{98}{3}+...+\dfrac{1}{100}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)

\(=\dfrac{\left(\dfrac{99}{2}+1\right)+\left(\dfrac{98}{3}+1\right)+...+\left(\dfrac{1}{100}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)

\(=\dfrac{\dfrac{101}{2}+\dfrac{101}{3}+...+\dfrac{101}{100}+\dfrac{101}{101}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)

\(=\dfrac{101\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)

\(=101-2\)

\(=99\)

22 tháng 1 2019

@Luân Đào

8 tháng 1 2018

\(a.\left\{{}\begin{matrix}4\dfrac{1}{x}+\dfrac{1}{y}=12\\\dfrac{1}{x}+\dfrac{1}{y}=-3\end{matrix}\right.\) (1)

ĐK xác định : x≠0 ; y≠0

Đặt ẩn phụ : a = \(\dfrac{1}{x}\) ; b = \(\dfrac{1}{y}\)

Thay vào (1) ta được :

\(\left\{{}\begin{matrix}4a+b=12\\a+b=-3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}3a=15\\a+b=-3\end{matrix}\right.< =>\left\{{}\begin{matrix}a=5\\b=-8\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{8}\end{matrix}\right.\)

Vậy S = {(\(\dfrac{1}{5};-\dfrac{1}{8}\))}

\(b.\left\{{}\begin{matrix}5\dfrac{1}{x}+2\dfrac{1}{y}=6\\2\dfrac{1}{x}-\dfrac{1}{y}=3\end{matrix}\right.\) (2)

ĐK xác định : x≠0 ; y≠0

Đặt ẩn phụ : a = 1/x ; b = 1/y

Thay vào (2) ta được : \(\left\{{}\begin{matrix}5a+2b=6\\2a-b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}5a+2b=6\\4a-2b=6\end{matrix}\right.< =>\left\{{}\begin{matrix}9a=12\\2a-b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=\dfrac{4}{3}\\b=-\dfrac{1}{3}\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-3\end{matrix}\right.\)

Vậy S = {(\(\dfrac{3}{4};-3\) )}

c) \(\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.\)

ĐK xác định : x≠0 ; y ≠0

Áp dụng quy tác cộng đại số ta có :

\(\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\3\dfrac{1}{x}-3\dfrac{1}{y}=15\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-3\dfrac{1}{y}=-13\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{3}{13}\\x=\dfrac{3}{28}\end{matrix}\right.\)

Vậy S = {(\(\dfrac{3}{28};\dfrac{3}{13}\))}

d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.\)

ĐK xác định : x≠0 ; y≠0

áp dụng quy tắc cộng đại số ta có :

\(\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.< =>\left\{{}\begin{matrix}2\dfrac{1}{x}-8\dfrac{1}{y}=10\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-5\dfrac{1}{y}=9\\\dfrac{1}{x}-4\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-\dfrac{5}{9}\\x=-\dfrac{5}{11}\end{matrix}\right.\)

Vậy S = {(\(-\dfrac{5}{11};-\dfrac{5}{9}\))}

e) ĐK xác định x≠0 ; y≠0

\(\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\6\dfrac{1}{x}-\dfrac{1}{y}=2\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\18\dfrac{1}{x}-3\dfrac{1}{y}=6\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-17\dfrac{1}{x}=-2\\\dfrac{1}{x}-3\dfrac{1}{y}=4\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=\dfrac{17}{2}\\y=-\dfrac{17}{22}\end{matrix}\right.\)

Vậy S={(\(\dfrac{17}{2};-\dfrac{17}{22}\))}

18 tháng 7 2017

2 = 1.2 => \(\dfrac{1}{2}\) = \(\dfrac{1}{1.2}\) = 1 - \(\dfrac{1}{2}\)

TT \(\dfrac{1}{6}=\dfrac{1}{2}-\dfrac{1}{3}\)

.................

=> VT = 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{\sqrt{2012-x}+2012}{\sqrt{2012-x}+2013}\)

Đặt \(\sqrt{2012-x}+2012=y\)

=> 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{y}{y+1}\)

=> \(\dfrac{x}{x+1}\) = \(\dfrac{y}{y+1}\)

=> x = y

<=> x = \(\sqrt{2012-x}+2012\)

<=> 2012 - x + \(\sqrt{2012-x}\) = 0

<=> \(\sqrt{2012-x}=0\)

<=> x = 2012