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Ta có:
A= 1+2-3-4+5+6-7-8+...-2011-2012+2013+2014
= (1+2-3-4)+(5+6-7-8)+...(2009+2010-2011-2012)+(2013+2014)
Ta thấy từ 1 đến 2012 có: \(x = {2012-1 \over 1}\)+1=2012(số)
Ta nhóm các số hạng kia trong tổng A và bớt đi tổng 2013+2014, mỗi nhóm là 4 số hạng liên tiếp
=> Có số nhóm là: 2012:4=503(nhóm)
Ta lại có:
A= (1+2-3-4)+(5+6-7-8)+...(2009+2010-2011-2012)+(2013+2014)
=(-4)+(-4)+...+(-4)+(2013+2014)
(503 số hạng -4)
=(-4).503+(2013+2014)
=(-2012)+4027
=2015
Vậy A=2015
Ta có : 1+2-3-4+5+6-7-8+...-2011-2012+2013+2014
=(1+2)+(-3-4+5+6)+(-7-8+9+10)+...+(-2011-2012+2013+2014)
=3+(4+4+...+4)(có 503 số 4)
=3+4*503
=3+2012
=2015
1) 5 + (-4) = 1
2) (-8) + 2 = -6
3) 8 + (-2) = 6
4) 11 + (-3) = 8
5) (-11) + 2 = -9
6) (-7) + 3 = -4
7) (-5) + 5 = 0
8) 11 + (-12) = -1
9) (-18) + 20 = 2
10) (15) + (-12) = 3
11) (-17) + 17 = 0
12) 16 + (-2) = 14
13) (30) + (-14) = 16
14) (-19) + 20 = 1
15) (-18) + 15 = -3
16) (10) + (-6) = 4
17) (-28) + 14 = -14
18) 15 + (-30) = -15
19) (15) + (-4) = 11
20) (-21) + 11 = -10
21) 8 + (-22) = -14
22) (-15) + 4 = -11
23) (-3) + 2 = -1
24) 17 + (-14) = 3
25) 17 + (-14) = 3
\(a.\frac{7}{4}+\frac{5}{6}\div5-\frac{3}{8}\left(-30\right)^2=\frac{7}{4}+\frac{1}{6}-\frac{675}{2}\)
\(=\frac{23}{12}-\frac{675}{2}\)
\(=-\frac{4027}{12}\)
\(b.\frac{4}{7}+\frac{3}{7}\div3-\frac{3}{8}\left(-2\right)^3=\frac{4}{7}+\frac{1}{7}-\left(-3\right)\)
\(=\frac{4}{7}+\frac{1}{7}+3\)
\(=\frac{5}{7}+3\)
\(=\frac{26}{7}\)
K cho mk nhé thks bn nhìu!
CHÚC BẠN HỌC TỐT
1: \(=-\dfrac{3}{4}+3-\dfrac{1}{4}-\dfrac{5}{4}+\dfrac{9}{2}=-\dfrac{9}{4}+3+\dfrac{9}{2}=\dfrac{-9+12+18}{4}=\dfrac{21}{4}\)
2: \(=\dfrac{-5}{7}\left(\dfrac{13}{19}+\dfrac{6}{19}\right)+\dfrac{5}{7}=\dfrac{-5}{7}+\dfrac{5}{7}=0\)
3: \(=\dfrac{2}{5}\left(1-\dfrac{8}{3}-\dfrac{5}{8}\right)=\dfrac{2}{5}\cdot\dfrac{-55}{24}=\dfrac{-11}{12}\)
\(\left(1\right)\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=-\dfrac{7}{12}.\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=-\dfrac{7}{12}.6+\dfrac{1}{2}=-3.\)
\(\left(2\right)\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4-\dfrac{2}{5}=\dfrac{-5}{3}-\dfrac{2}{5}=\dfrac{-31}{15}.\)
\(\left(3\right)\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=1-1-\dfrac{3}{2}=-\dfrac{3}{2}.\)
1. \(\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=\dfrac{-7}{12}\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=\dfrac{-7}{12}.\dfrac{6}{1}+\dfrac{1}{2}=\dfrac{-7}{2}+\dfrac{1}{2}=\dfrac{-6}{2}=-3\)2.
\(\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4+\dfrac{-2}{5}=\dfrac{-5}{3}+\dfrac{-2}{5}=\dfrac{-31}{15}\)
3.
\(\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=\dfrac{-4}{10}+\dfrac{3}{10}-\dfrac{6}{10}+\dfrac{7}{10}-\dfrac{15}{10}=\dfrac{-15}{10}=\dfrac{-3}{2}\)