Tính nhanh
a/ 2 12 . 13 + 2 12 . 65 2 10 . 104 + 3 10 . 11 + 3 10 . 5 3 9 . 2 4
b/ 1 2 .4 + 1 4 .6 + 1 6 .8 + . .. + 1 98 .100
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a. \(A=\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,75-2,2+\dfrac{11}{7}+\dfrac{11}{13}}=\dfrac{3\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)
Vậy \(A=\dfrac{3}{11}\)
b. \(B=\dfrac{2^{12}\cdot13+2^{12}\cdot65}{2^{10}\cdot104}+\dfrac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\dfrac{2^{12}\left(13+65\right)}{2^{10}\cdot104}+\dfrac{3^{10}\left(11+5\right)}{3^9\cdot2^4}=\dfrac{2^{12}\cdot78}{2^{10}\cdot104}+\dfrac{3^{10}\cdot16}{3^9\cdot16}=\dfrac{2^2\cdot3}{1\cdot4}+3=\dfrac{12}{4}+3=3+3=6\)
Vậy \(B=6\)
a) \(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(=\frac{2^{10}.\left(13.4+65.4\right)}{2^{10}.104}+\frac{3^9.\left(3.11+3.5\right)}{3^9.16}\)
\(=\frac{312}{104}+\frac{48}{16}\)
=3+3=6
b) \(\frac{1.5.6+2.10.12+4.20.24+9.45.54}{1.3.5+2.6.10+4.12.20+9.27.45}\)
\(=\frac{1.5.6\left(1+2.2.2+4.4.4+9.9.9\right)}{1.3.5\left(1+2.2.2+4.4.4+9.9.9\right)}\)
\(=\frac{1.5.6}{1.3.5}\)
\(=2\)
c) 1+2-3-4+5+6-7-8+...+2009+2010-2011-2012+2013
Nhận xét:Giá trị tuyệt đối của hai số liền nhau hơn kém nhau 1 đơn vị
=> Tổng trên có 2013-1+1=2013(Số hạng)
Nhóm 4 số vào một nhóm, ta được 2013:4=503 nhóm (thừa 1 số)
=>1+2-3-4+5+6-7-8+...+2009+2010-2011-2012+2013
=1+(2-3-4+5)+(6-7-8+9)+...+(2010-2011-2012+2013)
=1+0+0+...+0 (có 503 số 0)
=1+0.503
=1+0
=1
b, \(3737.43-4343.37=\left(37.101\right).43-\left(43.101\right).37=0\)
suy ra B = 0
c, \(D=\frac{2^{12}\left(13+65\right)}{2^{10}.104}+\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{2^{12}.78}{2^{10}.104}+\frac{3^{10}.16}{3^9.2^4}\)
\(=\frac{2^{12}.2.39}{2^{10}.2^3.13}+\frac{3^{10}.2^4}{3^9.2^4}=\frac{39}{13}+3=6\)
2) 
3) 
4) 
A=(2+3+...+13)-(1+2+...+12)=2+3+...+13-1-2-...-12=(13-1)+(2-2)+(3-3)+...+(12-12)=12
\(\dfrac{11}{12}\times\dfrac{12}{13}\times\dfrac{13}{14}:\dfrac{22}{28}\)
\(=\dfrac{11}{12}\times\dfrac{12}{13}\times\dfrac{13}{14}\times\dfrac{14}{11}\)
\(=\dfrac{11\times12\times13\times14}{12\times13\times14\times11}\)
\(=1\)
=================
\(3\times\dfrac{7}{10}+\dfrac{7}{10}\times5+2\times\dfrac{7}{10}\)
\(=\dfrac{7}{10}\times\left(3+5+2\right)\)
\(=\dfrac{7}{10}\times10=7\)
\(\dfrac{11}{12}\times\dfrac{12}{13}\times\dfrac{13}{14}:\dfrac{22}{28}\)
\(=\dfrac{11}{12}\times\dfrac{12}{13}\times\dfrac{13}{14}:\dfrac{11}{14}\)
\(=\dfrac{11}{12}\times\dfrac{12}{13}\times\dfrac{13}{14}\times\dfrac{14}{11}\)
\(=\dfrac{11\times12\times13\times14}{12\times13\times11\times14}\)
\(=1\)
\(3\times\dfrac{7}{10}+\dfrac{7}{10}\times5+2\times\dfrac{7}{10}\)
\(=\dfrac{7}{10}\times\left(3+5+2\right)\)
\(=\dfrac{7}{10}\times10\)
\(=7\)