nhân 1/2 cả 2 vế PT:

bạn tự làm típ nha

\(\frac{x}{6}+\frac{x}{10}+\frac{x}{15}+\frac{x}{21}+\frac{x}{28}+\frac{x}{36}=\frac{8}{9}\)

\(\Leftrightarrow\frac{x}{12}+\frac{x}{20}+\frac{x}{30}+\frac{x}{42}+\frac{x}{56}+\frac{x}{72}=\frac{8}{18}=\frac{4}{9}\)

\(\Leftrightarrow\frac{x}{3.4}+\frac{x}{4.5}+\frac{x}{5.6}+\frac{x}{6.7}+\frac{x}{7.8}+\frac{x}{8.9}=\frac{4}{9}\)

\(\Leftrightarrow\frac{x}{3}-\frac{x}{4}+\frac{x}{4}-\frac{x}{5}+\frac{x}{5}-\frac{x}{6}+...+\frac{x}{8}-\frac{x}{9}=\frac{4}{9}\)

\(\Leftrightarrow\frac{x}{3}-\frac{x}{9}=\frac{4}{9}\Leftrightarrow\frac{2x}{9}=\frac{4}{9}\Leftrightarrow x=2\)

Mấy con DOG ơi ! Ra đây tao cho ăn CỨT đồng loại nè !

bạn đợi mình tí 1+1=2

\(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)

\(\Rightarrow3B=1+\frac{1}{3}+...+\frac{1}{3^{2004}}\)

\(\Rightarrow3B-B=\left(1+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+...+\frac{1}{3^{2005}}\right)\)

\(\Rightarrow2B=1-\frac{1}{3^{2005}}< 1\)

\(\Rightarrow2B< 1\)

\(\Rightarrow B< \frac{1}{2}\left(đpcm\right)\)

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\)

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)

\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)

\(2B=1-\frac{1}{3^{2005}}\)

\(B=\frac{1-\frac{1}{3^{2005}}}{2}\)

\(B=\frac{1}{2}-\frac{1}{\frac{3^{2005}}{2}}\)

Vi  \(\frac{1}{2}-\frac{1}{\frac{3^{2005}}{2}}< \frac{1}{2}\)

\(\Rightarrow B< \frac{1}{2}\left(dpcm\right)\)

= 4/7 + 1/6 - 0,375 . 4

= 4/7 + 1/6 - 3/2

= -16/21

Học tốt nhé bn!

4/7+5/6:5-0,375.(-2)²

=4/7+1/6-3/8.4

=4/7+1/6-3/2

=31/42-3/2

=-16/21

#học tốt#

giúp mình nhanh nha mọi người 

Giúp cái dcm nhà mày

\(A=\left\{4;5;6\right\}\)

\(B=\left\{4;6;9;11\right\}\)

a. \(A=\left\{4,5,6\right\}\)

b. \(B=\left\{4,6,9,11\right\}\)

Học tốt nhé !

Ta có : \(n^2+4=n^2-4+8\) \(=\left(n-2\right)\left(n+2\right)+8\)

Để \(n+4⋮n+2\) thì \(8⋮n+2\)

\(\Rightarrow n+2\inƯ\left(8\right)\) . Vì n là số tự nhiên nên \(n+2\ge2\)

\(\Rightarrow n+2\in\left\{2;4;8\right\}\Rightarrow n\in\left\{0;2;6\right\}\)

Vậy => \(n^2+4⋮n+2\)

Ko ghi đề

A = (1+2-3-4)+(5+6-7-8)+...+(401+402-403-404)+405+406

A = (-2)+(-2)+...+(-2)+811  (có 101 số -2)

A = (-2) . 101 + 811

A = -202 + 811

=> A = 609

Vậy A = 609 

Nhớ đúng mk nha bn :)