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1 tháng 7 2016

4/1.5 + 4/5.9 + ... + 4/97.101 = 2x+5/101

=> 1 - 1/5 + 1/5 - 1/9 + ... + 1/97 - 1/101 = 2x+5/101

=> 1 - 1/101 = 2x+5/101

=> 100/101 = 2x+5/101

=> 2x + 5 = 100

=> 2x = 100 - 5 = 95

=> x = 95/2

Vậy x = 95/2

Ủng hộ mk nha ♡_♡☆_☆

1 tháng 7 2016

\(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{97.101}=\frac{2x+5}{101}\)

\(4-\frac{4}{5}+\frac{4}{5}-\frac{5}{9}+...+\frac{4}{97}-\frac{4}{101}=\frac{2x+5}{101}\)

\(4-\frac{4}{101}=\frac{2x+5}{101}\)

\(\frac{400}{101}=\frac{2x+5}{101}\)

\(\Rightarrow2x+5=400\)

\(\Rightarrow x=197,5\)

Vậy \(x=197,5\)

a, \(\frac{1}{1.4}\)+\(\frac{1}{4.7}\)+......+\(\frac{1}{97.100}\)= |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{1}{3}\) ( \(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+.......+\(\frac{3}{97.100}\))= |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{1}{3}\) ( 1  - \(\frac{1}{4}\)\(\frac{1}{4}\)-\(\frac{1}{7}\)+......+\(\frac{1}{97}\)-\(\frac{1}{100}\)) = |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{1}{3}\) ( 1-\(\frac{1}{100}\)) = |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{1}{3}\) . \(\frac{99}{100}\) = |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{33}{100}\) = |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{x}{3}\)\(\orbr{\begin{cases}\frac{33}{100}\\\frac{-33}{100}\end{cases}}\)

Với \(\frac{x}{3}\) = \(\frac{33}{100}\)

\(\Rightarrow\)100x= 33.3

 \(\Rightarrow\)100x=99

\(\Rightarrow\)x=\(\frac{99}{100}\)

Với \(\frac{x}{3}\)=\(\frac{-33}{100}\)

\(\Rightarrow\)100x=-33.3

\(\Rightarrow\)100x=-99

\(\Rightarrow\)x=\(\frac{-99}{100}\)

Vậy x=\(\orbr{\begin{cases}\frac{99}{100}\\\frac{-99}{100}\end{cases}}\)

b, \(\frac{4}{1.5}\)\(\frac{4}{5.9}\)+......+ \(\frac{4}{97.101}\)= |\(\frac{5x-4}{101}\)|

\(\Rightarrow\)1-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{9}\)+......+\(\frac{1}{97}\)-\(\frac{1}{101}\)= |\(\frac{5x-4}{101}\)|

\(\Rightarrow\)1-\(\frac{1}{101}\)= |\(\frac{5x-4}{101}\)

\(\Rightarrow\) \(\frac{100}{101}\)= |\(\frac{5x-4}{101}\)|

\(\Rightarrow\)\(\frac{5x-4}{101}\) =\(\orbr{\begin{cases}\frac{100}{101}\\\frac{-100}{101}\end{cases}}\)

Với \(\frac{5x-4}{101}\) =\(\frac{100}{101}\)

\(\Rightarrow\)(5x-4).101=100.101

\(\Rightarrow\)505x-404=10100

\(\Rightarrow\)505x=10504

\(\Rightarrow\)x=\(\frac{104}{5}\)

Với \(\frac{5x-4}{101}\)=\(\frac{-100}{101}\)

\(\Rightarrow\)(5x-4). 101=-100.101

\(\Rightarrow\)505x-404=-10100

\(\Rightarrow\)505x=-9696

\(\Rightarrow\)x=\(\frac{-96}{5}\)

Vậy x=\(\orbr{\begin{cases}\frac{104}{5}\\\frac{-96}{5}\end{cases}}\)

3 tháng 1 2018

\(M=-\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{n\left(n+4\right)}\right)\\ =-\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{n}-\dfrac{1}{n+4}\right)\\ =-\left(1-\dfrac{1}{n+4}\right)\\ =-\left(\dfrac{n+3}{n+4}\right)\)

3 tháng 1 2018

\(-4x\left(x-5\right)-2x\left(8-2x\right)=-3\\ \Rightarrow-4x^2+20x-16x+4x^2=-3\\ \Rightarrow4x=-3\\ \Rightarrow x=-\dfrac{3}{4}\)

VT >0 => VP > 0 => x >0

số dấu  | |  là  (397 - 1): 4 + 1 = 100

\(\Rightarrow100x+\left(\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{397.401}\right)=101x\)

\(\Rightarrow x=\left(\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{397.401}\right)\)

\(\Rightarrow4x=1-\frac{1}{401}\)

\(x=\frac{100}{401}\)(tm)

5 tháng 8 2015

\(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{177.181}+\frac{4}{181.185}\)

\(=\left(\frac{1}{1}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{13}\right)+...+\left(\frac{1}{177}-\frac{1}{181}\right)+\left(\frac{1}{181}-\frac{1}{185}\right)\)

\(=\frac{1}{1}-\frac{1}{185}\)

\(=\frac{184}{185}\)

29 tháng 10 2015

Có dạng tổng quát như thế này nhé: 
\(\frac{k}{n\left(n+k\right)}=\frac{1}{n}-\frac{1}{k+n}\)

Trong trường hợp này là \(\frac{-4}{1.5}-\frac{4}{5.9}-...-\frac{4}{\left(n+4\right)n}=-\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{n}-\frac{1}{n+4}\right)\)

Đáp án là: \(\frac{1}{n+4}-1\)

30 tháng 10 2017

\(\left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.14}\right|+...+\left|x+\dfrac{1}{397.401}\right|\ge0\)

\(\Rightarrow101x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+\dfrac{1}{1.5}+x+\dfrac{1}{5.9}+...+x+\dfrac{1}{397.401}=101x\)

\(\Rightarrow101x+\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{397.401}\right)=x\)

\(\Rightarrow\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{397.401}\right)=x\)

\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+....+\dfrac{1}{397}-\dfrac{1}{401}\right)\)

\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)\)

\(\Rightarrow x=\dfrac{1}{4}.\dfrac{400}{401}\)

\(\Rightarrow x=\dfrac{100}{401}\)