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S1=1+(-2)+...+2001+(-2002)
Có:(2002-1):1+1=2002(số)
S1=(1+(-2))+...+(2001+(-2002))
S1=(-1)+...+(-1)
Có:2002:2=1001(số)
=>S1=(-1).1001
=>S1=-1001
nhóm âm vào âm.dương vào dương
hoặc nhóm số đầu với số cuối số 2 với số kế cuối
12-12+11+10-9+8-7+5-4+3+2-1
=-(12+12-11-10)-(9-8+7-5)+(4-3-2+1)
=(-3)-3+0
=(-6)
Chúc bn học tốt
a,3/7 +1/2-[-3]/70
=13/14 -[-3]/70
=65/70 -[-3]/70
=34/35
b,3/5+-1/25 -35/100
=14/25 - 35/100
=56 /100 -35/100
= 21/100
c ,5/12 -3/-16+3/4
=29/48 +3/4
= 29/48 +36/48
=65/48
d, 5/15 +4/-12 +1/7 -1/-6
= 1/3 +1/-3 +1/7-1/-6
=0+1/7-1/-6
= 13/42
A.\(\dfrac{3}{7}+\dfrac{1}{2}+\dfrac{\left(-3\right)}{70}\)
=\([\dfrac{30}{70}+\dfrac{\left(-3\right)}{70}]\)+\(\dfrac{1}{2}\)
= \(\dfrac{27}{70}+\dfrac{35}{70}\)
=\(\dfrac{62}{70}=\dfrac{31}{35}\)
B.\(\dfrac{3}{5}+\dfrac{-1}{25}-\dfrac{35}{100}\)
=\(\dfrac{60}{100}+\dfrac{-4}{100}-\dfrac{35}{100}\)
=\(\dfrac{60+\left(-4\right)-35}{100}\)
=\(\dfrac{21}{100}\)
C.\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}\)
=\(\left(\dfrac{5}{12}+\dfrac{9}{12}\right)-\dfrac{-3}{16}\)
=\(\dfrac{7}{6}-\dfrac{-3}{16}\)
=\(\dfrac{56}{48}-\dfrac{-9}{48}\)=\(\dfrac{65}{48}\)
D.\(\dfrac{5}{15}+\dfrac{4}{-12}+\dfrac{1}{7}-\dfrac{1}{-6}\)
=\(\dfrac{1}{3}+\dfrac{1}{7}+(\dfrac{-2}{6}-\dfrac{-1}{6})\)
=\(\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{-1}{6}\)
=\((\dfrac{1}{3}+\dfrac{-1}{6})+\dfrac{1}{7}\)
=\(\dfrac{1}{6}+\dfrac{1}{7}\)=\(\dfrac{13}{42}\)
\(a,\frac{21}{36}.\frac{5}{2}-\frac{7}{12}.\frac{2}{7}+\left(2018-2019\right)^0\)
=\(\frac{7}{12}.\frac{5}{2}-\frac{7}{12}.\frac{2}{7}+\left(-1\right)\)
= \(\frac{7}{12}.\left(\frac{5}{2}+\frac{2}{7}\right)+\left(-1\right)\)
=\(\frac{7}{12}.\frac{39}{14}+\left(-1\right)\)
=\(\frac{13}{8}+\left(-1\right)\)
= \(\frac{5}{8}\)
\(b,-12\frac{1}{3}-\frac{5}{7}+7\frac{1}{3}+1\frac{5}{7}+1^{2019}\)
=\(-\frac{37}{3}+\frac{-5}{7}+\frac{22}{3}+\frac{12}{7}+1\)
=\(\left(\frac{-37+22}{3}\right)+\left(\frac{-5+12}{7}\right)+=1\)
= \(-5+1+1\)
=\(-3\)
1) \(\frac{7}{3}+\frac{1}{2}-\left(-\frac{3}{70}\right)=\frac{7}{3}+\frac{1}{2}+\frac{3}{70}=\frac{490}{210}+\frac{105}{210}+\frac{9}{210}=\frac{604}{210}=\frac{302}{105}\)
2) \(\frac{5}{12}-\left(\frac{3}{-16}\right)+\frac{3}{4}=\frac{5}{12}+\frac{3}{16}+\frac{3}{4}=\frac{20}{48}+\frac{9}{48}+\frac{36}{48}=\frac{65}{48}\)