0,55

-10\(\frac15\) + 7\(\frac12\) + 3\(\frac14\)

= \(\frac{-51}{5}\) + \(\frac{15}{2}\) + \(\frac{13}{4}\)

= \(\frac{-204}{20}\) + \(\frac{150}{20}\) + \(\frac{65}{20}\)

= \(\frac{-54}{20}\) + \(\frac{65}{20}\)

= \(\frac{11}{20}\)

Đặt \(S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\)

\(\Rightarrow2S=1+\frac{1}{2}+...+\frac{1}{2^{18}}\)

\(\Rightarrow2S-S=\left(1+\frac{1}{2}+...+\frac{1}{2^{18}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)\)

\(\Rightarrow S=1-\frac{1}{2^{19}}\)

Đặt S = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{19}}\)

=> 2S = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{18}}\)

2S - S = ( \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{18}}\)) - ( \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{19}}\))

S = 1 - \(\frac{1}{2^{19}}\)

\(\frac{49}{50}nha\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2106}\)

\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{2015}-\frac{1}{2016}\right)\)

\(A=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)

\(B=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2014.2016}=\frac{1}{4}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1007.1008}\right)\)

=> \(B=\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{1008}\right)=\frac{1}{4}.\frac{1007}{1008}\)

=> \(B=\frac{1007}{4032}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\)

mà \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}>\frac{5}{12}\)

\(0.25.30.2.\frac{1}{3}=0\)

\(\left|\frac{-2}{5}.x+\frac{1}{5}\right|-\frac{1}{4}=\frac{3}{4}\)

           \(\left|\frac{-2}{5}.x+\frac{1}{5}\right|=\frac{3}{4}+\frac{1}{4}\)

             \(\left|\frac{-2}{5}.x+\frac{1}{5}\right|=1\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{-2}{5}.x+\frac{1}{5}=1\\\frac{-2}{5}.x+\frac{1}{5}=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

đúng thì ủng hộ tớ nha

=5

=-2

CHÚC BẠN HỌC GIỎI

TK MÌNH NHÉ

\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2015}\right)\)

\(=\left(\frac{2}{2}-\frac{1}{2}\right).\left(\frac{3}{3}-\frac{1}{3}\right).\left(\frac{4}{4}-\frac{1}{4}\right)...\left(\frac{2015}{2015}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2014}{2015}\)

\(=\frac{1}{2015}\)

10ⁿ -1 luôn có dạng 99....

10²ⁿ -1 cũng luôn có dạng 99....

vì vậy 10²ⁿ -1 cũng sẽ chia hết cho 13

tick mình nha an nguyễn

ta co: \(10^{2n}-1=10^{2n}+10^n-10^n-1\)

                                   \(=\left(10^{2n}-10^n\right)+\left(10^n-1\right)\)                                                                                                               \(=10^n\left(10^n-1\right)+\left(10^n-1\right)\)

Vi \(10^n-1\) chia het cho 13 suy ra \(10^n\left(10^n-1\right)+\left(10^n-1\right)\)chia het cho 13

hay \(10^{2n}-1\) chia het cho 13

hay so du cua \(10^{2n}-1\) khi chia cho 13 la 0

Minh chac chan 100%

tick cho minh nha **********