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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1-2+3-4+...+97-98+99\)
\(=\left(1-2\right)+\left(3-4\right)+...+\left(97-98\right)+99\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+103\)
Có \(\left[\left(98-1\right):1+1\right]:2=49\)(cặp số \(\left(-1\right)\))
\(=49.\left(-1\right)+99=-49+99=50\)
\(B=\left(1-4\right)+\left(7-10\right)+...+\left(97-100\right)+103\)
\(=\left(-3\right)+\left(-3\right)+...+\left(-3\right)+103\)
Có \(\left[\left(100-1\right):3+1\right]:2=17\)(cặp số \(\left(-1\right)\))
\(=17.\left(-3\right)+103=\left(-51\right)+103=52\)
=) \(\frac{A}{B}=\frac{50}{52}=\frac{25}{26}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có B=..................=5(1+5) + 5^3(1+5)+...+5^7(1+5) = 6*5 + 6*5^3 + 6*5^7 = 6(5+...+5^7) chia hết cho 6
chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{3}.\frac{-9}{10}.\frac{-6}{13}.\frac{-13}{36}=\frac{-3}{10}.\frac{-1}{6}=\frac{1}{20}\)
\(B=\frac{4}{19}\left(\frac{-5}{12}+\frac{-7}{12}\right)-\frac{40}{57}=\frac{-4}{19}-\frac{40}{57}=\frac{-52}{57}\)
2 câu còn lại tự làm
\(A=\frac{1}{3}.\frac{-6}{13}.\frac{-9}{10}.\frac{-13}{36}\)
\(A=\frac{1}{1}.\frac{-2}{13}.\frac{-9}{10}.\frac{-13}{36}\)
\(A=\frac{-2}{13}.\frac{-9}{10}.\frac{-13}{36}\)
\(A=\frac{-1}{13}.\frac{-9}{5}.\frac{-13}{36}\)
\(A=\frac{-1}{13}.\frac{-1}{5}.\frac{-13}{4}\)
\(A=\frac{-13}{260}=\frac{-1}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{38}>\dfrac{1}{40}>\dfrac{1}{42}>...>\dfrac{1}{50}\)
=>\(\dfrac{1}{38}+\dfrac{1}{40}+\dfrac{1}{42}+\dfrac{1}{44}+\dfrac{1}{46}+\dfrac{1}{48}+\dfrac{1}{50}< 7\cdot\dfrac{1}{38}=\dfrac{7}{38}< 1\)
Vậy tổng trên bé hơn 1
A=-1-3-5-...-2017
=-(1+3+5+...+2017)
Xét tổng B=1+3+5+...+2017
Tổng B có:(2017-1):2+1=1009(số hạng)
Tổng B=\(\dfrac{\left(2017+1\right)\cdot1009}{2}=1009\cdot1009=1018081\)
=>A=-B=-1018081
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1)\frac{1}{2}x-\frac{3}{5}=\frac{-4}{5}\)
\(\Rightarrow\frac{1}{2}x=\frac{-4}{5}+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}x=\frac{-1}{5}\)
\(\Rightarrow x=\frac{-1}{5}:\frac{1}{2}=\frac{-1}{5}\cdot\frac{2}{1}=\frac{-2}{5}\)
\(\Leftrightarrow x=\frac{-2}{5}\)
\(2)3\frac{1}{5}-2\frac{1}{3}x=-1\frac{3}{5}+1\frac{7}{10}\)
\(\Rightarrow\frac{16}{5}-\frac{7}{3}x=-\frac{8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{16}{5}-\frac{-8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{16}{5}+\frac{8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{24}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{48}{10}+\frac{17}{10}\)
Đến đây tìm được rồi nhé
3,4, áp dụng bài 1,2 rồi làm :v
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\frac{-3}{4}+\frac{3}{7}+-\frac{1}{4}+\frac{4}{9}+\frac{4}{7}\)
\(=\left(\frac{-3}{4}+-\frac{1}{4}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{4}{9}\)
\(=-1+1+\frac{4}{9}\)
\(=\frac{4}{9}\)
\(\frac{-3}{4}+\frac{3}{7}+\frac{-1}{4}+\frac{4}{9}+\frac{4}{7}\)
\(=\left(\frac{-3}{4}+\frac{-1}{4}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{4}{9}\)
\(=\left(-1\right)+1+\frac{4}{9}\)
\(=0+\frac{4}{9}\)
\(=\frac{4}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(A=\dfrac{10^7-8+13}{10^7-8}=1+\dfrac{13}{10^7-8}\)
\(B=\dfrac{10^8-7+13}{10^8-7}=1+\dfrac{13}{10^8-7}\)
mà \(10^7-8< 10^8-7\)
nên A>B
c: \(\dfrac{1}{10}A=\dfrac{10^{1992}+1}{10^{1992}+10}=1-\dfrac{9}{10^{1992}+10}\)
\(\dfrac{1}{10}B=\dfrac{10^{1993}+1}{10^{1993}+10}=1-\dfrac{9}{10^{1993}+10}\)
mà \(\dfrac{9}{10^{1992}+10}>\dfrac{9}{10^{1993}+10}\)
nên A<B
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)
Vậy A < B
b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)
\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)
Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)
c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:
\(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)
Vậy A < B
![](https://rs.olm.vn/images/avt/0.png?1311)
1. a) \(\frac{-2}{7}+\frac{15}{23}+\frac{\left(-15\right)}{17}+\frac{4}{19}+\frac{8}{23}\)
\(=\left(\frac{-2}{7}+\frac{-5}{7}\right)+\left(\frac{15}{23}+\frac{8}{23}\right)+\frac{4}{19}\)
\(=\left(-1\right)+1+\frac{4}{19}\)
\(=0+\frac{4}{19}=\frac{4}{19}\)
b) \(\frac{7}{19}\cdot\frac{8}{11}+\frac{7}{19}\cdot\frac{3}{11}+\frac{12}{19}\)
\(=\frac{7}{19}\cdot\left(\frac{8}{11}+\frac{3}{11}\right)+\frac{12}{19}\)
\(=\frac{7}{19}\cdot1+\frac{12}{19}\)
\(=\frac{7}{19}+\frac{12}{19}=\frac{19}{19}=1\)
2. a) \(\frac{1}{3}+\frac{\left(-2\right)}{16}-\frac{7}{14}\)
\(=\frac{5}{24}-\frac{1}{2}\)
\(=-\frac{7}{24}\)
b) \(11\frac{3}{13}-2\frac{4}{7}+5\frac{3}{13}\)
\(=\left(11-2+5\right)+\frac{3}{13}-\frac{4}{7}+\frac{3}{13}\)
\(=14+\left(-\frac{10}{91}\right)\)
\(=-14\frac{10}{91}\)
c) \(0,7\cdot2\frac{2}{3}\cdot20\cdot0,375\cdot\frac{5}{28}\)
\(=\frac{7}{10}\cdot\frac{8}{3}\cdot20\cdot\frac{3}{8}\cdot\frac{5}{28}\)
\(=\left(\frac{7}{10}\cdot\frac{5}{28}\right)\cdot\left(\frac{8}{3}\cdot\frac{3}{8}\right)\cdot20\)
\(=\frac{1}{8}\cdot1\cdot20\)
\(=\frac{20}{8}=\frac{5}{2}\)
d) \(\frac{6}{7}+\frac{5}{7}:5-\frac{8}{9}\)
\(=\frac{6}{7}+\frac{1}{7}-\frac{8}{9}\)
\(=1-\frac{8}{9}\)
\(=\frac{1}{9}\)
~Học tốt~
a) Số lượng số hạng là:
`(2023-1):2+1=1012` (số hạng)
Số lượng cặp là:
`1012:2=506` (cặp)
`P=1-3+5-7+...+2021-2023`
`=(1-3)+(5-7)+...+(2021-2023)`
`=(-2)+(-2)+...+(-2)`
`=(-2)*506`
`=-1012`
b) Số lượng số hạng là:
`(103-1):3+1=34` (số hạng)
Số lượng cặp là:
`34:2=17(cặp)
`Q=1-4+7-10+...+100-103`
`=(1-4)+(7-10)+...+(100-103)`
`=(-3)+(-3)+...+(-3)`
`=(-3)*17`
`=-51`
P=1-3+5-7+...+2021-2023
=(1-3) + (5-7)+...+(2021+2023) (có 506 nhóm)
=(-2)+...+(-2) có 506 số hạng
=(-2). 506 = -1012
Kết quả: -1012