Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{\left(12:13\right)+\left(12:131\right)-\left(12:1313\right)+\left(12:13131\right)}{\left(15:13\right)+\left(15:131\right)-\left(15:1313\right)+\left(15:13131\right)}\\ =\dfrac{12:\left(13+131-1313+13131\right)}{15:\left(13+131-1313+13131\right)}=\dfrac{12}{15}=\dfrac{4}{5}\)
12.( -131-25)+131.( 12+25)
= -12.131-12.25+131.12+131.25
= -12.25+131.25
= 25.( 131-12)
= 25.119
= 2975
#Châu's ngốc
\(\frac{\frac{5}{17}+\frac{5}{49}+\frac{5}{131}}{\frac{7}{17}+\frac{7}{49}+\frac{7}{131}}\)
\(=\frac{5.\left(\frac{1}{17}+\frac{1}{49}+\frac{1}{131}\right)}{7.\left(\frac{1}{17}+\frac{1}{49}+\frac{1}{131}\right)}\)
\(=\frac{5}{7}\)
=\(\frac{5.\left(\frac{1}{17}+\frac{1}{49}+\frac{1}{131}\right)}{7.\left(\frac{1}{17}+\frac{1}{49}+\frac{1}{131}\right)}\)
=\(\frac{5}{7}\)
a. 131.1961+1000/131.1961+1961-961
=131.1961+1000/131.1961+1000=1
b(2-4-6+8)+(10-12-14+16)+...+(2002-2004-2006+2008)=0
=>B=\(\frac{12.\left(\frac{1}{13}+\frac{1}{1313}+\frac{1}{131}+\frac{1}{1313}\right)}{15.\left(\frac{1}{13}+\frac{1}{1313}+\frac{1}{131}+\frac{1}{1313}\right)}\)
=>B=\(\frac{12}{15}\)
=>B=\(\frac{4}{5}\)
B = \(\frac{12.\left(\frac{1}{13}+\frac{1}{1313}+\frac{1}{131}-\frac{1}{1313}\right)}{15.\left(\frac{1}{13}+\frac{1}{131}-\frac{1}{1313}+\frac{1}{1313}\right)}\)
=\(\frac{12.\left(\frac{1}{13}+\frac{1}{131}\right)}{15.\left(\frac{1}{13}+\frac{1}{131}\right)}\)
=\(\frac{12}{15}=\frac{4}{5}\)
Vậy B = 4/5.
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
\(A=\frac{12}{3.5}+\frac{12}{5.7}+...+\frac{12}{2013.2015}\)
\(2A=\frac{24}{3.5}+\frac{24}{5.7}+...+\frac{24}{2013.2015}\)
\(2A=\frac{24}{3}-\frac{24}{5}+\frac{24}{5}-\frac{24}{7}+...+\frac{24}{2013}-\frac{24}{2015}\)
\(2A=8-\frac{24}{2015}\)
\(2A=\frac{8}{1}-\frac{24}{2015}\)
\(2A=\frac{16120}{2015}-\frac{24}{2015}\)
\(2A=\frac{16096}{2015}\)
\(=>A=\frac{16096}{2015}:2\)
\(=>A=\frac{16096}{4030}\)
a, 2009
b, -24