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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,\(\dfrac{3.5.7.11.13.37-10101.55}{1212120+40404}\)
\(=\dfrac{55\left(3.7.13.37-10101\right)}{1212120+40404}\)
\(=\dfrac{55.0}{1212120+40404}=0\)
\(\frac{5+55+555+5555}{9+99+999+9999}=\frac{5\cdot\left(1+11+111+1111\right)}{9\cdot\left(1+11+111+1111\right)}=\frac{5}{9}\)
A=(9/1999+99/999+999/9999).(1/5-1/4+1/20)
A=(9/1999+99/999+999/9999).(-1/20+1/20)
A=(9/1999+99/999+999/9999).0
A=0
Vì mọi số nhân vs 0 thì đều = 0 kể cả phân số
mk nhanh nhất ủng hộ nha
\(A=\left(\frac{9}{1999}+\frac{99}{999}+\frac{999}{9999}\right)\cdot0\)
A=0
Bài 1.
b) \(\frac{5+55+555+5555}{9+99+999+9999}\)
= \(\frac{5\left(1+11+111+1111\right)}{9\left(1+11+111+1111\right)}=\frac{5}{9}\)
c) \(39,2\cdot27+39,2\cdot43+78,4\cdot15\)
= \(39,2\cdot27+39,2\cdot43+39,2\cdot2\cdot15\)
= \(39,2\left(27+43+30\right)=39,2\cdot100=3920\)
d) \(\frac{4}{17}\cdot\frac{3}{11}+\frac{8}{11}\cdot\frac{4}{17}-\frac{4}{17}\)
= \(\frac{4}{17}\left(\frac{3}{11}+\frac{8}{11}-1\right)=\frac{4}{17}\cdot0=0\)
Bài 2.
a) \(\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{57\cdot59}\)
= \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{57}-\frac{1}{59}\)
= \(\frac{1}{5}-\frac{1}{59}=\frac{54}{295}\)
b) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)-\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)\)
= \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\)
= \(\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
c) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2012}\right)\)
= \(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2011}{2012}=\frac{1}{2012}\)
ta có 5 + 55 +555 + 5555 + 55555 +555555 +5555555 +55555555 +555555555 + 5555555555
rồi bấm máy tính là ra
ủng hộ tui với nhé
\(A=\frac{5+55+555+5555}{9+99+999+9999}=\frac{5.\left(1+11+111+1111\right)}{9.\left(1+11+111+1111\right)}=\frac{5}{9}\)
Tớ làm luôn
A=\(\frac{5.\left(1+11+111+1111\right)}{9.\left(1+11+111+1111\right)}\)
A=\(\frac{5}{9}\)