(1+10) +(2+9) +....+ (5+6) + (10+1) +(9+2) +........+ (6+5)

= 11 x 10 = 110

Giải

 (1+9)+(2+8)...(6+4)+(7+3)...(9+1)+10+10

10*11=110

a/A=1+2+4+8+...+1024

2A=2+4+8+16+....+2048

2A-A=(2+4+8+16+....+2048)-(1+2+4+8+...+1024)

A=2048-1

A=2047

VẬY A=2047

b/B=1+5+25+125+....+15625

5B=5+25+125+625+....+78125

5B-B=(5+25+125+625+....+78125)-(1+5+25+125+....+15625)

4B=78125-1

4B=78124

B=78124:4

B=19531

VẬY B =19531

C=1/1.2+1/2.3+1/3.4+...+1/2015.2016

C=1-1/2+1/2-1/3+1/3-1/4+...+1/2015-1/2016

=1-1/2016

=2015/2016

VẬY C=2015/2016

D/=10/1.3+10/3.5+10/5.7+....+10/2013.2015

=5(2/1.3+2/3.5+2/5.7+...+2/2013.2015)

=5(1-1/3+1/3-1/5+1/5-1/7+..+1/2013-1/2015)

=5(1-1/2015)

=5.2014/2015

=2014/403

VẬY D=2014/403

a, A = 1 + 2 + 4 + 8 +...+ 1024

   \(A=1+2+2^2+2^3+....+2^{10}\)

   \(2A=2+2^2+2^3+....+2^{10}+2^{11}\)

   \(A=1+2+2^2+2^3+....+2^{10}\)

  \(A=2^{11}-1=2047\)

b, B = 1 + 5 + 25 + 125 + ... + 15625

   \(B=1+5+5^2+5^3+....+5^6\)

   \(3B=5+5^2+5^3+....+5^6+5^7\)

   \(B=1+5+5^2+5^3+....+5^6\)

  \(2B=5^7-1\Rightarrow B=\frac{5^7-1}{2}=39062\)

d, D = 10 / 1 . 3 + 10 / 3 . 5 + 10 / 5 . 7 + ... + 10 / 2013 . 2015

\(D=\frac{10}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)

\(D=\frac{10}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(D=\frac{10}{2}.\left(1-\frac{1}{2015}\right)=5.\frac{2014}{2015}=\frac{2014}{403}\)

Câu c thì tương tự

\(2\dfrac{1}{3}.3=\dfrac{7}{3}.3=7.\\ \left(\dfrac{2}{5}-\dfrac{3}{4}\right)-\dfrac{2}{5}=\dfrac{2}{5}-\dfrac{3}{4}-\dfrac{2}{5}=-\dfrac{3}{4}.\\ \dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}.\\ =\dfrac{-10}{11}\left(\dfrac{4}{7}+\dfrac{3}{7}-1\right).\\ =\dfrac{-10}{11}.\left(1-1\right)=0.\)

1) 2\(\dfrac{1}{3}\).3=\(\dfrac{7}{3}\).3=7.

2) (2/5 -3/4) -2/5 = 2/5 -3/4 -2/5 = -3/4.

3) \(\dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}=\dfrac{1}{11}\left(-\dfrac{40}{7}-\dfrac{30}{7}+21\right)=\dfrac{1}{11}.\left(-10+21\right)=1\).

\(a,3456731-19994=3436737\)

\(b,3\times31\times16+2\times24\times42+4\times27\times12\)

\(=\left(3\times16\right)\times31+\left(2\times24\right)\times42+\left(4\times12\right)\times27\)

\(=48\times31+48\times42+48\times27\)

\(=48\times\left(31+42+27\right)\)

\(=48\times100\)

\(=4800\)

\(c,\left(3^{10}+3^{12}\right):3^{10}\)

\(=3^{10}:3^{10}+3^{12}:3^{10}\)

\(=1+3^2\)

\(=1+9\)

\(=10\)

\(d,\left(2^{13}-3.2^{10}\right):2^{10}+\left(5^{10}-5^9\right):5^9\)

\(=2^{13}:2^{10}-3.2^{10}:2^{10}+5^{10}:5^9-5^9:5^9\)

\(=2^3-3+5-1\)

\(=8-3+5-1\)

\(=9\)

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)

\(2A=1-\frac{1}{3^{100}}\)

\(A=\frac{1-\frac{1}{3^{100}}}{2}\)

\(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)

\(B=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)

\(3B=\frac{5.3}{4.7}+\frac{5.3}{7.10}+\frac{5.3}{10.13}+...+\frac{5.3}{25.28}\)

\(3B=5\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)

\(3B=5\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)

\(3B=5\left(\frac{1}{4}-\frac{1}{28}\right)\)

\(3B=5\cdot\frac{3}{14}=\frac{15}{14}\)

\(B=\frac{15}{14}:3=\frac{5}{14}\)

a) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)

\(2A=1-\frac{1}{3^{100}}\)

\(\Rightarrow A=\frac{1-\frac{1}{3^{100}}}{2}\)

b)  \(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)

\(B=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)

\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{5}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+\frac{5}{3}.\left(\frac{1}{10}-\frac{1}{13}\right)+...+\frac{5}{3}.\left(\frac{1}{25}-\frac{1}{28}\right)\)

\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)

\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)\)

\(B=\frac{5}{3}.\frac{3}{14}\)

\(\Rightarrow B=\frac{5}{14}\)