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\(a,A=\dfrac{201220122012}{201320132013}=\dfrac{201220122012:100010001}{201320132013:100010001}=\dfrac{2012}{2013}\)
\(b,B=\dfrac{3.5.7.11.13.37-10101}{1212120+40404}=\dfrac{5.11.10101-10101}{10101.120+10101.4}=\dfrac{10101.\left(5.11-1\right)}{10101.\left(120+4\right)}=\dfrac{54}{124}=\dfrac{27}{62}\)
\(B=\dfrac{40404}{70707}+\dfrac{244\times395-151}{244+395\times243}+\dfrac{1\times3\times5+2\times6\times10+4\times12\times20+7\times21\times35}{1\times5\times7+2\times10\times14+4\times20\times28+7\times35\times49}\\ =\dfrac{4}{7}+\dfrac{243\times395+395-151}{244+395\times243}+\dfrac{1\times3\times5\left(1+2+4+7\right)}{1\times5\times7\left(1+2+4+7\right)}\\ =\dfrac{4}{7}+\dfrac{243\times395+244}{244+395\times243}+\dfrac{3}{7}\\ =\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+1\\ =1+1=2\)
A = \(\frac{3^4\left(5.79-1\right)}{2^2.3^2\left(5.79-1\right)}=\frac{9}{4}\); B = \(\frac{3.7.13.37\left(5.11-1\right)}{4.3.7.13.37\left(3.5-1\right)}=\frac{54}{4.14}=\frac{2.9}{4.2.7}=\frac{9}{28}\)
a, \(\dfrac{3.5.7.11.13.37-10101.55}{1212120+40404}\)
\(=\dfrac{55\left(3.7.13.37-10101\right)}{1212120+40404}\)
\(=\dfrac{55.0}{1212120+40404}=0\)
b, \(\dfrac{5+55+555+5555}{9+99+999+9999}\)
\(=\dfrac{5.\left(1+11+111+1111\right)}{9.\left(1+11+111+1111\right)}=\dfrac{5}{9}\)
Chúc bạn học tốt!!!
a,\(\dfrac{3.5.7.11.13.37-10101.55}{1212120+40404}\)
\(=\dfrac{55\left(3.7.13.37-10101\right)}{1212120+40404}\)
\(=\dfrac{55.0}{1212120+40404}=0\)
\(\Rightarrow\)A = \(\dfrac{\left(3.7.13.37\right).\left(5.11\right)-10101.1}{40404.30+40404.1}\)
\(\Rightarrow\)A = \(\dfrac{10101.55-10101.1}{40404.\left(30+1\right)}\)
\(\Rightarrow\)A =\(\dfrac{10101.\left(55-1\right)}{10101.\left(4.30\right)}\)
\(\Rightarrow\)A= \(\dfrac{10101.54}{10101.124}\)
chia cả tử và mẫu cho 10101 thì
\(\Rightarrow\)A= \(\dfrac{54}{124}\)
\(\Rightarrow\)A= \(\dfrac{27}{62}\)
Vậy A =\(\dfrac{27}{62}\)
à bạn ơi mình đánh nhầm số ở mẫu phải là 10101.(4.31) chứ ko phải 10101.(4.30) đâu
\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{197}-\dfrac{1}{199}\)
\(A=\dfrac{1}{3}-\dfrac{1}{199}\)
\(A=\dfrac{199}{597}-\dfrac{3}{597}=\dfrac{196}{597}\)
\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{197.199}\)
\(A=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+...+\dfrac{199-197}{197.199}\)
\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{197}-\dfrac{1}{199}\)
\(A=\dfrac{1}{3}-\dfrac{1}{999}\)
\(A=\dfrac{196}{697}\)
\(B=1+2+4+8+16+...+512+1024\)
\(2B=2+4+8+32+...+1024+2048\)
\(B=\left(2+4+8+...+2048\right)-\left(1+2+4+...+1024\right)\)
\(B=2048-1\)
\(B=2047\)
\(M=\dfrac{201220122012}{201320132013}=\dfrac{2012}{2013}\)
\(N=\dfrac{1326395265}{1836547290}=\dfrac{13}{18}\)