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18 tháng 9 2017

\(\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)..............\left(1-\dfrac{2}{99.100}\right)\)

\(=\left(\dfrac{6}{2.3}-\dfrac{2}{2.3}\right).\left(\dfrac{12}{3.4}-\dfrac{2}{3.4}\right)..............\left(\dfrac{9900}{99.100}-\dfrac{2}{99.100}\right)\)

\(=\dfrac{4}{2.3}.\dfrac{10}{3.4}..........................\dfrac{9898}{99.100}\)

\(=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}..............\dfrac{98.101}{99.100}\)

\(=\dfrac{1.2.3.....98}{2.3......99}.\dfrac{4.5.6.....101}{3.4.......100}\)

\(=\dfrac{1}{99}.\dfrac{101}{3}=\dfrac{101}{297}\)

18 tháng 9 2017

bn ơi ! còn phần tìm x thì sao

giải giúp mk đi khocroi

7 tháng 3 2017

22 là thế nào đấy bạn?

7 tháng 3 2017

2 mủ 2 đấy bn

22 tháng 3 2017

a)

ta có:

\(\left\{{}\begin{matrix}\dfrac{b-a}{b-a}=1..\forall a\ne b\\\dfrac{b-a}{a.b}=\dfrac{1}{a}-\dfrac{1}{b}..\forall a,b\ne0\end{matrix}\right.\)(*)

\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+..+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(\left\{{}\begin{matrix}a=3n-1\\b=3n+2\end{matrix}\right.\)\(\Rightarrow b-a=3..\forall n\)

Thay (*) vào dãy A

\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-....+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)

\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)=\dfrac{1}{3}\left(\dfrac{3n+2-2}{2.\left(3n+2\right)}\right)=\dfrac{n}{6n+4}=VP\rightarrow dpcm\)

B) tương tự

25 tháng 3 2017

Cảm ơn bạn

9 tháng 4 2017

nhầm đề

9 tháng 4 2017

phá ngoặc kiểu j vậy bn

8 tháng 8 2017

\(4\dfrac{1}{3}\cdot\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\\ \dfrac{13}{3}\cdot\dfrac{-1}{3}\le x\le\dfrac{2}{3}\cdot\dfrac{-11}{12}\\ \dfrac{-13}{9}\le x\le\dfrac{-11}{18}\\ \dfrac{-26}{18}\le x\le\dfrac{-11}{18}\\ \Rightarrow x=-1\)

8 tháng 8 2017

điều kiện là gì?

\(x\in N,x\in Z,x\in Q?\)

19 tháng 9 2017

\(\)\(\left|x-1,5\right|+\left|2,5-x\right|=0\)

Với mọi \(x\in R\) thì:

\(\left\{{}\begin{matrix}\left|x-1,5\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix} \left|x-1,5\right|=0\\ \left|2,5-x\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\)

Khi đó không tồn tại giá trị x

\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{6}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\sqrt{\dfrac{1}{6}}\\x+\dfrac{1}{2}=-\sqrt{\dfrac{1}{6}}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}+\sqrt{\dfrac{1}{6}}\\x=\dfrac{1}{2}-\sqrt{\dfrac{1}{6}}\end{matrix}\right.\)

19 tháng 9 2017

\(\sqrt{\dfrac{1}{6}=?}\)

mk ko hiểu Linh Nguyễn

mk chưa hk đến căn

2 tháng 4 2017

\(=\dfrac{2}{2}\).(\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+...+\(\dfrac{2}{x.\left(x+1\right)}\))

=2.(\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+...+\(\dfrac{2}{x.\left(x+1\right)}\))

=2.(\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+...+\(\dfrac{1}{x.\left(x+1\right)}\))

=2.[(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\))+(\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\))+(\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\))+...+(\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\))

=2.[\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\)]

2.[(\(\dfrac{1}{3}\)-\(\dfrac{1}{3}\))+(\(\dfrac{1}{4}\)-\(\dfrac{1}{4}\))+...+(\(\dfrac{1}{x}\)-\(\dfrac{1}{x}\))+(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))]

=2.[0+0+...+0+(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))]

=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))

=2.(\(\dfrac{1.x+1-1.2}{2.x+1}\))

=2.(\(\dfrac{x+1-2}{2x}\))=2.\(\dfrac{x-1}{2x}\)=\(\dfrac{2.\left(x-1\right)}{2x}\)=\(\dfrac{2x-2}{2x}\)

\(\dfrac{2x-2}{2x}\)=\(\dfrac{2014}{2016}\)\(\Rightarrow\)(2x-2).2016=2014.2x=4032x-4032=4028x

\(\Rightarrow\)4032x-4028x=4x=4032\(\Rightarrow\)x=4032:4=1008

2 tháng 4 2017

Đặt A=\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x.\left(x+1\right)}\)

\(A=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}\)

\(A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x.\left(x+1\right)}\)

16 tháng 4 2017

ừ Vy Nguyễn, mik làm nè:

e, \(\dfrac{-2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}.\)

\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-2}{3}-\dfrac{3}{2}.\)

\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-4}{6}+\dfrac{-9}{6}.\)

\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-13}{6}.\)

\(2x-5=\dfrac{-13}{6}:\dfrac{1}{3}.\)

\(2x-5=\dfrac{-13}{6}.3.\)

\(2x-5=\dfrac{-13}{2}.\)

\(2x=\dfrac{-13}{2}+5.\)

\(2x=\dfrac{-13}{2}+\dfrac{10}{2}.\)

\(2x=\dfrac{-3}{2}.\)

\(x=\dfrac{-3}{2}:2.\)

\(x=\dfrac{-3}{2.2}=\dfrac{-3}{4}.\)

g, \(\dfrac{2}{5}x+\dfrac{1}{2}=\dfrac{-3}{4}.\)

\(\dfrac{2}{5}x=\dfrac{-3}{4}-\dfrac{1}{2}.\)

\(\dfrac{2}{5}x=\dfrac{-3}{4}+\dfrac{-2}{4}.\)

\(\dfrac{2}{5}x=\dfrac{-5}{4}.\)

\(x=\dfrac{-5}{4}:\dfrac{2}{5}.\)

\(x=\dfrac{-5}{4}.\dfrac{5}{2}.\)

\(x=\dfrac{-25}{8}.\)

h, \(\left(2x-2\dfrac{4}{5}\right):3\dfrac{1}{8}=1\dfrac{3}{5}.\)

\(\left(2x-2\dfrac{4}{5}\right)=\dfrac{8}{5}.\dfrac{25}{8}.\)

\(\left(2x-2\dfrac{4}{5}\right)=5.\)

\(2x=5+2\dfrac{4}{5}.\)

\(2x=7\dfrac{4}{5}.\)

\(x=7\dfrac{4}{5}:2.\)

\(x=\dfrac{39}{10}.\)

(còn tiếp ở phần sau!!!)

16 tháng 4 2017

Tiếp:

i, \(3,2x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):3\dfrac{2}{3}=\dfrac{7}{20}.\)

\(\dfrac{16}{5}x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right)=\dfrac{7}{20}.\dfrac{11}{3}.\)

\(\dfrac{16}{5}x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right)=\dfrac{77}{60}.\)

\(\dfrac{16}{5}x-\left(\dfrac{12}{15}+\dfrac{10}{15}\right)=\dfrac{77}{60}.\)

\(\dfrac{16}{5}x-\dfrac{22}{15}=\dfrac{77}{60}.\)

\(\dfrac{16}{5}x=\dfrac{77}{60}+\dfrac{22}{15}.\)

\(\dfrac{16}{5}x=\dfrac{77}{60}+\dfrac{88}{60}.\)

\(\dfrac{16}{5}x=\dfrac{165}{60}=\dfrac{11}{4}.\)

\(x=\dfrac{11}{4}:\dfrac{16}{5}.\)

\(x=\dfrac{11}{4}.\dfrac{5}{16}=\dfrac{55}{64}.\)

k, \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}.\)

\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right).\)

\(\left(\dfrac{3x}{7}+1\right)=\dfrac{1}{7}.\)

\(\dfrac{3x}{7}=\dfrac{1}{7}-1.\)

\(\dfrac{3x}{7}=\dfrac{1}{7}-\dfrac{7}{7}.\)

\(\dfrac{3x}{7}=\dfrac{-6}{7}.\)

\(\Rightarrow3x=-6.\)

\(\Rightarrow x=-6:3=-2.\)

~ Chúc bn học tốt!!! ~

Bài mik đúng thì nhớ tik mik nha!!!