5 12 + 1 3 2 + 3 4 − 5 6 2 = 5 12 + 4 12 2 + 9 12 − 10 12 2 = 3 4 2 + − 1 12 2 = 9 16 + 1 144 = 41 72

\(512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-......-\frac{512}{2^{10}}\)

\(=512.\left(1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-....-\frac{1}{2^{10}}\right)\)

Đặt \(A=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-....-\frac{1}{2^{10}}\)

\(=>2A=2-1-\frac{1}{2}-\frac{1}{2^2}-....-\frac{1}{2^9}\)

\(=>2A-A=\left(2-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^9}\right)-\left(1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-....-\frac{1}{2^{10}}\right)\)

\(=>A=2-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^9}-1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{10}}\)

\(=>A=2-1-1+\frac{1}{2^{10}}=\frac{1}{2^{10}}\)

\(=>512-\frac{512}{2}-\frac{512}{2^2}-...-\frac{512}{2^{10}}=512.\frac{1}{2^{10}}=\frac{512}{2^{10}}=\frac{1}{2}\)

\(=512\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

\(=512\left(1-\frac{1}{2^{10}}\right)=512.\frac{1023}{1024}=\frac{1023}{512}\)

    17 \(\times\) ( \(\dfrac{1313}{5151}\) + \(\dfrac{1111}{3434}\)) : \(\dfrac{117}{512}\)

= 17 \(\times\) ( \(\dfrac{1313:101}{5151:101}\) + \(\dfrac{1111:101}{3434:101}\)) : \(\dfrac{117}{512}\)

= 17 \(\times\) ( \(\dfrac{13}{51}\) + \(\dfrac{11}{34}\)): \(\dfrac{117}{512}\)

= 17 \(\times\) \(\dfrac{59}{102}\) \(\times\) \(\dfrac{512}{117}\)

= \(\dfrac{1003}{102}\) \(\times\) \(\dfrac{512}{117}\)

=            \(\dfrac{15104}{351}\)

\(M=512-\frac{512}{2}-\frac{512}{2^2}-...-\frac{512}{2^{10}}\)

\(M=512-512.\left(\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\right)\)

Đặt\(S=\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)

=> \(\frac{1}{2}S=\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{11}}\)

=> \(\frac{1}{2}S-S=-\frac{1}{2}S=\frac{1}{2^{11}}-\frac{1}{2}\)

=> \(S=\left(\frac{1}{2^{11}}-\frac{1}{2}\right):-\frac{1}{2}\)