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\(S=1+\left(-2\right)+\left(-3\right)+4+5+\left(-6\right)+\left(-7\right)+8+9+...+2001\)
\(\Rightarrow S=\left[1+\left(-2\right)+\left(-3\right)+4\right]+\left[5+\left(-6\right)+\left(-7\right)+8\right]+...+2001\)
\(\Rightarrow S=0+0+0+...+0=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\left(+15\right)+\left(+17\right)=15+17=32\)
2) \(\left(-3\right)+\left(-7\right)=-3-7=-\left(3+7\right)=-10\)
3) \(\left(-25\right)+\left(+4\right)=-25+4=-\left(25-4\right)=-21\)
4) \(\left(-6\right)+\left(-54\right)=-6-54=-\left(6+54\right)=-60\)
5) \(\left(-15\right)+20=20-15=5\)
6) \(\left(-5\right)+8+7+5\)
\(=\left(-5+5\right)+\left(8+7\right)\)
\(=15\)
7) \(\left(-8\right)+\left(-11\right)+\left(-2\right)\)
\(=\left[\left(-8\right)+\left(-2\right)\right]+\left(-11\right)\)
\(=\left(-10\right)+\left(-11\right)\)
\(=-21\)
8) \(15+\left(-5\right)+\left(-14\right)+\left(-16\right)\)
\(=\left[15+\left(-5\right)\right]+\left[\left(-14\right)+\left(-16\right)\right]\)
\(=10+\left(-30\right)\)
\(=-20\)
9) \(\left(-20\right)+\left(-14\right)+3+\left(-86\right)\)
\(=\left[\left(-20\right)+3\right]+\left[\left(-14\right)+\left(-86\right)\right]\)
\(=\left(-17\right)+\left(-100\right)\)
\(=-117\)
10) \(\left(-136\right)+123+\left(-264\right)+\left(-83\right)+240\)
\(=\left[\left(-136\right)+\left(-264\right)\right]+\left[123+\left(-83\right)\right]+240\)
\(=\left(-400\right)+40+240\)
\(=\left(-360\right)+240\)
\(=-120\)
11) \(\left(-596\right)+2001+1999+\left(-404+189\right)\)
\(=\left(-596\right)+2001+1999-404+189\)
\(=\left[\left(-596\right)-404\right]+\left(2001+189\right)+1999\)
\(=\left(-1000\right)+2190+1999\)
\(=1190+1999\)
\(=3189\)
12) \(314+\left(-153\right)+64+121+\left(-247\right)+218\)
\(=\left(314+64+121\right)+\left[\left(-153\right)+\left(-247\right)\right]+218\)
\(=\left(378+121\right)+\left(-400\right)+218\)
\(=499-400+218\)
\(=99+218\)
\(=317\)
\(\text{#}Toru\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sai đề, mình nghĩ là - 2007
E có: (2007 - 1) : 2 + 1 = 1004 (số)
Khi nhóm 2 số thành 1 nhóm thì có: 1004 : 2 = 502 (nhóm)
Ta có:
E = (1 - 3) + (5 - 7) + (9 - 11) +...+ (2005 - 2007) (502 nhóm)
E = (-2) + (-2) + (-2) +...+ (-2) (502 số)
E = (-2).502 = -1004
![](https://rs.olm.vn/images/avt/0.png?1311)
1001 nhé
nhóm 2 số và 1 nhóm có kq = -1.có 1000 nhóm và thừa số 2001
-1000+2001=1001
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1 - 3 + 5 - 7 + 9 - 11 + ... + 2001 - 2003
= (1 - 3) + (5 - 7) + (9 - 11) + ... + (2001 - 2003) (1002 cặp)
= -2 + (-2) + (-2) + ... + (-2) (501 số -2)
= -2.501 = -1002
\(1-3+5-7+9-11+...+2001-2003\)
\(=\left(1-3\right)+\left(5-7\right)+...+\left(2001-2003\right)\)
\(=\left(-2\right)+\left(-2\right)+\left(-2\right)+...+\left(-2\right)\)
\(=\left(-2\right)\times501\)(vì có 501 cặp số bằng -2)
\(=-1002\)