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14 tháng 7 2018

=1/1-1/4+1/4-1/7+....+1/n-1/n+3

=1-1/n+3

=n+2/n+3

14 tháng 7 2018

Ta có : 

3/ 1.4 + 3/ 4.7 + 3/ 7.10 + ... + 3/ n( n + 1 )

= 1 - 1/4 + 1/4 - 1/7 + ... + 1/ n - 1/ n + 3 .

= 1 - 1/ n+3 .

= n+3 - 1 / n+3 

= n+2 / n+3 .

21 tháng 2 2016

\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{n}-\frac{1}{n+3}\)

     \(=1-\frac{1}{n+3}<1\)

7 tháng 9 2016

bạn ơi như là cô giáo cho đề sai rồi kết quả phải là \(\frac{375}{376}\)thì mới giải được

12 tháng 8 2017

Ta có:

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)

\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{125}{376}\)

\(\Rightarrow\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{125}{376}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{x+3}=\frac{125}{376}:\frac{1}{3}=\frac{375}{376}\)

\(\Rightarrow\frac{1}{x+3}=1-\frac{375}{376}=\frac{1}{376}\Leftrightarrow x+3=376\Leftrightarrow x=373\)

7 tháng 9 2016

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)

\(3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{125}{376}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{375}{376}\)

\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{375}{376}\)

\(1-\frac{1}{x+3}=\frac{375}{376}\)

\(\frac{x+2}{x+3}=\frac{375}{376}\)

=> x + 2 = 375

=> x = 375 - 2

=> x = 373

1 tháng 9 2019

a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)

\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)

\(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)

\(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)

\(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)

\(\frac{1}{x+1}=\frac{1}{18}\)

⇒ x + 1 = 18

⇒ x = 17

Vậy x = 17

b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)

\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)

\(1-\frac{1}{x+3}=\frac{147}{148}\)

\(\frac{1}{x+3}=1-\frac{147}{148}\)

\(\frac{1}{x+3}=\frac{1}{148}\)

⇒ x + 3 = 148

⇒ x = 145

Vậy x = 145

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

10 tháng 11 2016

Đặt vế trái phương trình là A

\(3A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}\)

\(3A=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+\frac{\left(x+3\right)-x}{x\left(x+3\right)}\)

\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\)

\(3A=1-\frac{1}{x+3}=\frac{x+2}{x+3}\Rightarrow A=\frac{x+2}{3\left(x+3\right)}\)

\(\Rightarrow\frac{x+2}{3\left(x+3\right)}=\frac{667}{2002}\Rightarrow2002\left(x+2\right)=3.667.\left(x+3\right)\)

\(\Leftrightarrow2002x+4004=2001x+6003\Leftrightarrow x=1999\)

1 tháng 10 2016

anh gửi chơi hay thật vậy