40 - 5 x 6 - 10 = 40 - 30 - 10 = 40 - (30 + 10) = 40 - 40 = 0

=200 

jk mik nhé

40-5.6-10=35.(-4)

             =-140

\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{3}{10}\)

Ta có: \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{3}{10}\)

\(\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{3}{10}\)

\(\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{3}{10}\)

\(\dfrac{1}{x+1}=\dfrac{1}{30}\)

\(\Rightarrow x+1=30\)

\(x=30-1\)

\(x=29\)

Vậy ...

\(=\frac{2^{30}.3^{18}.2.3.7-3.2^{30}.3^{20}}{2^8.5^2.2^{20}.3^{20}-2^{10}.2^{19}.3^{19}.5.7}\)

\(=\frac{2^{31}.3^{19}.7-2^{30}.3^{21}}{2^{28}.5^2.3^{20}-2^{29}.3^{19}.5.7}\)

\(=\frac{2^{30}.3^{19}\left(2.7-3^2\right)}{2^{28}.3^{19}\left(5^2.3-2.5.7\right)}\)

\(=\frac{2^2.5}{5}=\frac{20}{5}=4\)

\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + .....+\(\dfrac{1}{n.(n+1)}\) = \(\dfrac{3}{10}\)

\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\) +......+ \(\dfrac{1}{n}-\dfrac{1}{n+1}\) = \(\dfrac{3}{10}\)

\(\dfrac{1}{3}-\dfrac{1}{n+1}\) = \(\dfrac{3}{10}\)

         \(\dfrac{1}{n+1}\) = \(\dfrac{1}{3}-\dfrac{3}{10}\)

          \(\dfrac{1}{n+1}\) = \(\dfrac{1}{30}\)

           n + 1 = 30

           n = 30 - 1

           n = 29

Kết luận n = 29 là giá  trị thỏa mãn yêu cầu đề bài.

\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{3}{10}\)

\(\dfrac{1}{3}-\dfrac{1}{n+1}=\dfrac{3}{10}\)

\(\dfrac{-1}{\left(n+1\right)}=\dfrac{-1}{30}\)

\(-n-1=-30\)

-n = -29

n = 29