Tính tổng sau bằng phương pháp hợp lý nhất:
A=2/3*5+2/5*7+2/7*9+......+2/37*39
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KV
30 tháng 10 2023
Câu 9
a) 2/15 + (-3/5)
= 2/15 - 3/5
= 2/15 - 9/15
= -7/15
b) -5/7 + 3/7 . (-2/9)
= -5/7 - 2/21
= -15/21 - 2/21
= -17/21
c) 5/16 - (7/15 - 3/16) + 17/30
= 5/16 - 7/15 + 3/16 + 17/30
= (5/16 + 3/16) + (17/30 - 14/30)
= 1/2 + 1/10
= 5/10 + 1/10
= 6/10
= 3/5
NM
30 tháng 10 2023
\(a,\dfrac{2}{15}+\left(-\dfrac{3}{5}\right)\\ =\dfrac{2}{15}-\dfrac{3}{5}\\ =\dfrac{2}{15}-\dfrac{9}{15}\\ =-\dfrac{7}{15}\\ b,-\dfrac{5}{7}+\dfrac{3}{7}\times\left(-\dfrac{2}{9}\right)\\ =-\dfrac{5}{7}-\dfrac{2}{21}\\ =-\dfrac{15}{21}-\dfrac{2}{21}\\ =-\dfrac{17}{21}\\ c,\dfrac{5}{16}-\left(\dfrac{7}{15}-\dfrac{3}{16}\right)+\dfrac{17}{30}\\ =\dfrac{5}{16}-\dfrac{7}{15}+\dfrac{3}{16}+\dfrac{17}{30}\\ =\left(\dfrac{5}{16}+\dfrac{3}{16}\right)-\left(\dfrac{17}{15}+\dfrac{17}{30}\right)\\ =\dfrac{1}{2}-\dfrac{17}{10}\\ =\dfrac{5}{10}-\dfrac{17}{10}\\ =-\dfrac{6}{5}.\)
M
22 tháng 6 2019
1. Q=165+247+528+125+315
= 100+60+5+200+40+7+500+20+8+100+20+5+300+10+5
= ( 100+200+500+100+300) + ( 60+40+20+20+10 ) +( 5+7+8+5+5 )
= 1200+150+30
=1280
R=1234+5678+8+80
=1000+200+30+4+5000+600+70+8+8+80
= ( 1000+5000) + ( 200+600 ) + (30+70+80) + ( 4+8+8)
=6000+800+180+20
=7000
22 tháng 12 2015
a)A có: (26-12)/2+1=8(số hạng)
A=(26+12)*8/2=152
b)B=1+(-2)+3+(-4)+...+2003+(-2004)+2005=(-2+1)+(-4+3)+...+(-2004+2003)+2005=(-1)+(-1)+...+(-1)+2005=(-1)*1002+2005=-1002+2005=1003
haizzz, còn lại lười làm quá
O
9 tháng 5 2022
a) \(\dfrac{7}{30}+\dfrac{\left(-12\right)}{37}+\dfrac{23}{30}+\dfrac{\left(-25\right)}{37}=\left(\dfrac{7}{30}+\dfrac{23}{30}\right)+\left(\dfrac{-12}{37}+\dfrac{-25}{37}\right)=1+\left(-1\right)=0\)
b) \(\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}=\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)=-\dfrac{5}{11}\)
c) \(\dfrac{\left(-5\right)}{7}\cdot\dfrac{3}{13}-\dfrac{5}{7}\cdot\dfrac{10}{13}+1\dfrac{5}{7}=\dfrac{5}{7}\cdot\dfrac{-3}{13}-\dfrac{5}{7}\cdot\dfrac{10}{13}+\dfrac{12}{7}=\dfrac{5}{7}\cdot\left(\dfrac{-3}{13}-\dfrac{10}{13}\right)+\dfrac{12}{7}=\dfrac{5}{7}\cdot\left(-1\right)+\dfrac{12}{7}=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=\dfrac{7}{7}=1\)
\(B=\dfrac{2}{3.5}+\dfrac{2}{3.7}+...+\dfrac{2}{37.39}\)
\(B=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(B=\dfrac{1}{3}-\dfrac{1}{39}\)
\(B=\dfrac{12}{39}=\dfrac{4}{3}\)