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nếu tìm S
S5=1+4+7+...+79
=>5S=(79+1)[(79-1):3+1]:2
=>5S=80[78:3+1]:2
=>5S=80.27:2
=>5S=1080
=>S=216
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(S_1=1+2+3+4+......+999\)
\(\Rightarrow S_1=\dfrac{\left(999+1\right).\left[\left(999-1\right):1+1\right]}{2}\)
\(\Rightarrow S_1=\dfrac{1000.\left(998+1\right)}{2}\)
\(\Rightarrow S_1=\dfrac{1000.999}{2}\)
\(\Rightarrow S_1=\dfrac{999000}{2}\)
\(\Rightarrow S_1=499500\)
b) \(S_2=10+12+14+......+2010\)
\(\Rightarrow S_2=\dfrac{\left(2010+10\right).\left[\left(2010-10\right):2+1\right]}{2}\)
\(\Rightarrow S_2=\dfrac{2020.\left(2000:2+1\right)}{2}\)
\(\Rightarrow S_2=\dfrac{2020.\left(1000+1\right)}{2}\)
\(\Rightarrow S_2=\dfrac{2020.1001}{2}\)
\(\Rightarrow S_2=\dfrac{2022020}{2}\)
\(\Rightarrow S_2=1011010\)
c) \(S_3=21+23+25+.......1001\)
\(\Rightarrow S_3=\dfrac{\left(1001+21\right).\left[\left(1001-21\right):2+1\right]}{2}\)
\(\Rightarrow S_3=\dfrac{1022.\left(980:2+1\right)}{2}\)
\(\Rightarrow S_3=\dfrac{1022.\left(490+1\right)}{2}\)
\(\Rightarrow S_3=\dfrac{1022.491}{2}\)
\(\Rightarrow S_3=\dfrac{501802}{2}\)
\(\Rightarrow S_3=250901\)
d) \(S_5=1+4+7+......+79\)
\(\Rightarrow S_5=\dfrac{\left(79+1\right).\left[\left(79-1\right):3+1\right]}{2}\)
\(\Rightarrow S_5=\dfrac{80.\left(78:3+1\right)}{2}\)
\(\Rightarrow S_5=\dfrac{80.\left(26+1\right)}{2}\)
\(\Rightarrow S_5=\dfrac{80.27}{2}\)
\(\Rightarrow S_5=\dfrac{2160}{2}\)
\(\Rightarrow S_5=1080\)
e) \(S_7=15+25+35+45+......+115\)
\(\Rightarrow S_7=\dfrac{\left(115+15\right).\left[\left(115-15\right):10+1\right]}{2}\)
\(\Rightarrow S_7=\dfrac{130.\left(100:10+1\right)}{2}\)
\(\Rightarrow S_7=\dfrac{130.\left(10+1\right)}{2}\)
\(\Rightarrow S_7=\dfrac{130.11}{2}\)
\(\Rightarrow S_7=\dfrac{1430}{2}\)
\(\Rightarrow S_7=715\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,[(79+1).27]/2=1080
b,[(155+15).71]/2=6035
c,[(115+15).11]/2=715
d,[(24+126).103]/2=7725
tick nha
![](https://rs.olm.vn/images/avt/0.png?1311)
a,S1=1+2+3+...+999 (có 999 số hạng)=(1+999).999:2=499500
b,S2=10+12+14+...+2010 (có 1001 số hạng)=(10+2010).1001:2=1011010
c,S3=21+23+25+...+1001 (có 491 số hạng)=(21+1001).491:2=250901
d,S5=1+4+7+...+79 (có 27 số hạng)=(1+79).27:2=1080.
S1=1+2+3+...+999
Số các số hạng là:999-1:1+1=999
Tổng số các số hạng là:(1+999).999:2=499500
Phần tính tổng cậu nhớ chia theo khoảng cách nhé.
Tương tự các phần còn lại cũng thế chúc cậu làm bài tốt.
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2 . tìm x :
a ) 52x = 625
52x = 54
=> 2 . x = 4
x = 4 : 2
x = 2
b ) 9x-1 = 9
9x-1 = 91
=> x - 1 = 1
x = 1 + 1
x = 2
c ) 2x : 25 = 1
2x-5 = 1
=> x = 5 , vì 25-5 = 20 = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
tick cho mình nha
c) S3=(1001+21).491:2 = 250901
d) S5=(79+1).27:2=10
e) S6=(155+15).71:2=6035
f) S7= (115+15).21:2=1365
g) S4=(126+24).103:2=7725
S5=1+4+7+...+79 (có 27 số hạng)
=(1+79).27:2=1080.
@
S5=1+4+7+.............+79
S5=(1+79)*26:2=1040
\(\Rightarrow S=1040:5=208\)