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1, (1+3+5+....+2007+2009+2011)(125125.127-127127.125)
= (1+3+5+....+2007+2009+2011)(125.1001.127-127127127.125)
= (1+3+5+....+2007+2009+2011)(125.127127-127127.125)
= (1+3+5+....+2007+2009+2011) . 0
=0
còn bài 2 đề hình như thiếu nha bn
chúc bn học tốt!
125125 x 127 - 127127 x 125
= 1001 x 125 x 127 - 1001 x 127 x125
=0
k đi nha mn!
A=\(\frac{3}{9.6}+\frac{5}{2.7.9}+\frac{7}{6.7.7}+\frac{8}{3.7.29}\) = \(\frac{1}{18}+\frac{5}{7.18}+\frac{1}{6.7}+\frac{8}{3.7.29}\)
= \(\frac{7+5}{7.18}+\frac{1}{6.7}+\frac{8}{3.7.29}\)=\(\frac{2}{7.3}+\frac{1}{6.7}+\frac{8}{3.7.29}\)= \(\frac{1}{3.7}.\)(\(\frac{2}{1}+\frac{1}{2}+\frac{8}{29}\))=
=\(\frac{1}{21}.\)(\(\frac{116+29+16}{58}\))=\(\frac{1}{21}.\frac{161}{58}=\frac{161}{1218}\)
\(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)
\(=\left(6\frac{5}{7}-2\frac{5}{7}\right)-1\frac{3}{4}\)
\(=4-1\frac{3}{4}\)
\(=2\frac{1}{4}\)
\(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)
\(=\left(6\frac{5}{7}-2\frac{5}{7}\right)-1\frac{3}{4}\)
\(=4-\frac{7}{4}\)
\(=\frac{16}{4}-\frac{7}{4}\)
\(=\frac{9}{4}\)
= \(6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}=6\frac{5}{7}-2\frac{5}{7}\)\(-1\frac{3}{4}\)= \(4-1\frac{3}{4}\)=\(\frac{9}{4}\)
47/7 - (7/4 + 19/7) =47/7 -125/28 =gomen =tao xin lỗi chịu rùi = <3
Số số hạng : ( 49 - 1 ) : 2 +1 = 25
Tổng : ( 49 + 1 ) x 25 : 2 = 625
1+2-3-4+5+6-7-8+...+111-112+113+114+115
=1+(2-3-4+5)+(6-7-8+9)+....................................+(110-111-112+113)+114+115
=230
= 3/2 + 2/3 + 2/4 + 2/5
= (6/4 + 2/4) + (2/3 + 2/5)
= 8/4 + 16/ 15
= 46/15
=(1+3+5+7+...+2007).0
=0
=(1+3+5+7+........+2007).(125125.127-127127.125)
=(1+3+5+...........+2007). 0
=0
tk mk nha