\(\frac{2}{4}+\frac{2}{28}+\frac{2}{70}+\frac{2}{130}+\frac{2}{208}\)

= \(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}\)

= \(\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{13.16}\right)\)

= \(\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)

= \(\frac{2}{3}.\left(1-\frac{1}{16}\right)\)

= \(\frac{2}{3}.\frac{15}{16}\)

= \(\frac{5}{8}\)

2/4 + 2/28 + 2/70 + 2/130 + 2/208

= 2/1.4 + 2/4.7 + 1/7.10 + 1/10.13 + 1/13.16

= 2/3.(1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + 1/13 - 1/16)

= 2/3.(1 - 1/16)

= 2/3.15/16

= 5/8

P/s: "." là nhân nhé :V

=13/16

cách của mình ngắn gọn hơn cơ

a ) ( 3n - 2 )3n

b) n(n 1) ( n 2 )
 

pt đã cho có dạng \(\frac{1}{\left(x+1\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+10\right)}=\frac{4}{13}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+10}=\frac{4}{13}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{4}{13}\Leftrightarrow....\)

bạn tuấn mình thấy vậy nè

Gỉa sử cho x=1 ta thấy \(\frac{1}{1\times4}\ne\frac{1}{1}-\frac{1}{4}\)

Bạn bấm máy tính thử xem dấu bằng chỉ áp dụng với 2 số tự nhiên liên tiếp thôi còn cái này cách 3 lận

giải thích giúp mình với

à 

ý là nhân hai về pt cho 3 đúg kk 

mình hiểu rồi nha

cảm ơn nha

ta có x²+5x+4

=x²+x+4x+4

=(x²+x)+(4x+4)

=x(x+1)+4(x+1)

=(x+1)(x+4)

tương tự ta đc

x²+11x+28=(x+4)(x+7)

x²+17x+70=(x+7)(x+10)

x²+23x+130=(x+10)(x+13)

=>\(\dfrac{1}{\left(x+1\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+10\right)}+\dfrac{1}{\left(x+10\right)\left(x+13\right)}=\dfrac{4}{13}\)\(\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{3}{\left(x+4\right)\left(x+7\right)}+\dfrac{3}{\left(x+7\right)\left(x+10\right)}+\dfrac{3}{\left(x+10\right)\left(x+11\right)}=\dfrac{4}{13}\)=>\(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}+....+\dfrac{1}{x+13}=\dfrac{4}{13}\)

=>\(\dfrac{1}{x+1}-\dfrac{1}{x+13}=\dfrac{4}{13}\)

=>\(\dfrac{13\left(x+13\right)}{13\left(x+1\right)\left(x+13\right)}-\dfrac{13\left(x+1\right)}{13\left(x+1\right)\left(x+13\right)}=\dfrac{4\left(x+1\right)\left(x+13\right)}{13\left(x+1\right)\left(x+13\right)}\)

=> 13(x+13)-13(x+1)=4(x+1)(x+13)

=> 13[(x+13)-(x+1)]=(4x+4)(x+13)

=>13(x+13-x-1)=4x²+52x+4x+52

=13.12=4x²+56x+52

=>4x²+56x+52=156

=>4x²+56x-104=0

\(pt\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+10\right)}+\dfrac{1}{\left(x+10\right)\left(x+13\right)}=\dfrac{4}{13}\)

a) \(\dfrac{6}{13}:\left(\dfrac{1}{2}-x\right)=\dfrac{15}{39}\)

\(\dfrac{1}{2}-x=\dfrac{6}{13}:\dfrac{15}{39}\)

\(\dfrac{1}{2}-x=\dfrac{6}{5}\)

\(x=\dfrac{1}{2}-\dfrac{6}{5}\)

\(x=-\dfrac{7}{10}\)

b) \(3\times\left(\dfrac{x}{4}+\dfrac{x}{28}+\dfrac{x}{70}+\dfrac{x}{130}\right)=\dfrac{60}{13}\)

\(3\times x\times\left(\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}\right)=\dfrac{60}{13}\)

\(x\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{7\times13}\right)=\dfrac{60}{13}\)

\(x\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}\right)=\dfrac{60}{13}\)

\(x\times\left(1-\dfrac{1}{13}\right)=\dfrac{60}{13}\)

\(x\times\dfrac{12}{13}=\dfrac{60}{13}\)

\(x=\dfrac{60}{13}:\dfrac{12}{13}\)

\(x=5\)