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ZR
3 tháng 2 2016
\(\frac{1.2.3+2.4.6+4.8.12+7.14.21}{1.3.6+2.6.12+4.12.24+7.21.42}=\frac{1.2.3\left(1+2+4+7\right)}{1.3.6\left(1+2+4+7\right)}=\frac{1.2.3}{1.3.6}=\frac{2}{6}=\frac{1}{3}\)
NH
0
TQ
4
17 tháng 5 2017
\(\dfrac{1.2.3+2.4.6+4.8.12+7.14.21}{1.3.6+2.6.12+4.12.24+7.21.42}\)
\(=\dfrac{1.2.3\left(1+2.2.2+4.4.4+7.7.7\right)}{1.3.6\left(1+2.2.2+4.4.4+7.7.7\right)}\)
\(=\dfrac{1.2.3}{1.3.6}=\dfrac{1}{3}\)
TM
17 tháng 5 2017
Ta có:
\(\dfrac{1.2.3+2.4.6+4.8.12+7.14.21}{1.3.6+2.6.12+4.12.24+7.21.41}\)
=\(\dfrac{1.2.3\left(1+2^3+4^3+7^3\right)}{1.3.6\left(1+2^3+4^3+7^3\right)}\)= \(\dfrac{1.2.3}{1.3.6}\) = \(\dfrac{1}{3}\)
AT
17 tháng 5 2017
Ta có:
\(A=\dfrac{2002}{2000.16-1970}=\dfrac{2002}{\left(2002-2\right).16-1970}\)
\(=\dfrac{2001}{2002.16-2.16-2002}=\dfrac{2002}{2002.16-2002}\)
\(=\dfrac{2002}{2002.\left(16-1\right)}=\dfrac{1}{16-1}=\dfrac{1}{15}\)
\(\Rightarrow A=\dfrac{1}{15}\)
\(B=\dfrac{1.2.3+2.4.6+4.8.12+7.14.21}{1.3.6+2.6.12+4.12.24+7.21.42}\)
\(=\dfrac{1.2.3.\left(1+2.2.2+4.4.4+7.7.7\right)}{1.3.6.\left(1+2.2.2+4.4.4+7.7.7\right)}\)
\(=\dfrac{1.2.3}{1.3.6}=\dfrac{1}{3}\)
\(\Rightarrow B=\dfrac{1}{3}\)
Vì \(\dfrac{1}{3}>\dfrac{1}{15}\Rightarrow A< B\)
Vậy A < B
NT
18 tháng 5 2017
\(A=\dfrac{2002}{2000.16-1970}=\dfrac{2002}{2002.16-32-1970}=\dfrac{2002}{2002.15}=\dfrac{1}{15}\)
\(B=\dfrac{1.2.3\left(1.1.1+2.2.2+4.4.4+7.7.7\right)}{1.3.6\left(1.1.1+2.2.2+4.4.4+7.7.7\right)}=\dfrac{1}{3}=\dfrac{5}{15}>\dfrac{1}{15}=A\\ \Rightarrow B>A\)
Q
19 tháng 3 2017
\(\dfrac{2483-13}{4966-26}=\dfrac{2470}{4940}=\dfrac{2470:2470}{4940:2470}=\dfrac{1}{2}\)
\(\dfrac{2727-101}{7575+303}=\dfrac{2626}{7878}=\dfrac{2626:2626}{7878:2626}=\dfrac{1}{3}\)
\(\dfrac{2002}{2000.16-1970}=\dfrac{2002}{32000-1970}=\dfrac{2002}{30030}=\dfrac{2002:2002}{30030:2002}=\dfrac{1}{15}\)
\(\dfrac{1.2.3+2.4.6+4.8.12+7.4.21}{1.3.6+2.6.12+4.12.24+7.21.42}=\dfrac{6+48+384+588}{18+144+1152+6174}=\dfrac{54+972}{162+7326}\)
\(=\dfrac{1026}{7488}=\dfrac{1026:18}{7488:18}=\dfrac{57}{416}\)
Sau khi rút gọn ta có các phân số: \(\dfrac{1}{2};\dfrac{1}{3};\dfrac{1}{15};\dfrac{57}{416}\)
\(BCNN\left(2;3;15;416\right)=6240\)
\(\dfrac{1}{2}=\dfrac{1.3120}{2.3120}=\dfrac{3120}{6240}\)
\(\dfrac{1}{3}=\dfrac{1.2080}{3.2080}=\dfrac{2080}{6240}\)
\(\dfrac{1}{15}=\dfrac{1.416}{15.416}=\dfrac{416}{6240}\)
\(\dfrac{57}{416}=\dfrac{57.15}{416.15}=\dfrac{855}{6240}\)
\(B=\frac{1.2.3+2.4.6+4.8.12+7.14.21}{1.3.6+2.6.12+4.12.24+7.21.42}=\frac{1.2.3\left(1+2+4+7\right)}{1.3.6\left(1+2+4+7\right)}=\frac{1.2.3}{1.3.6}\frac{2}{6}=\frac{1}{3}\)