11+12-1
1+9+10-2-3
11+11+11-19
K
Khách
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VQ
10 tháng 8 2023
a, theo thứ tự tăng dần : -16/19;-14/19;-12/19;-11/19;-9/19;-3/19;-1/19
b, theo thứ tự tăng dần :-10/11;-10/9;-10/8;-10/7;-10/4;-10/3;-10/2
11 tháng 8 2023
a, theo thứ tự tăng dần : -16/19;-14/19;-12/19;-11/19;-9/19;-3/19;-1/19
b, theo thứ tự tăng dần :-10/11;-10/9;-10/8;-10/7;-10/4;-10/3;-10/2
29 tháng 5 2018
\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)
MA
4
24 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\dfrac{1}{2}-\dfrac{5}{6}+\dfrac{11}{33}-\dfrac{35}{40}\)
`=`\(\dfrac{1}{2}-\dfrac{5}{6}+\dfrac{1}{3}-\dfrac{7}{8}\)
`=`\(\dfrac{12}{24}-\dfrac{20}{24}+\dfrac{8}{24}-\dfrac{21}{24}\)
`= -21/24 = -7/8`
`b)`
\(\dfrac{2}{3}\cdot1\dfrac{3}{4}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)
`=`\(\dfrac{2}{3}\cdot\dfrac{7}{4}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)
`=`\(\dfrac{7}{6}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)
`=`\(\dfrac{5}{18}-\dfrac{17}{51}-\dfrac{1}{5}\)
`=`\(-\dfrac{1}{18}-\dfrac{1}{5}=-\dfrac{23}{90}\)
`c)`
\(\dfrac{1}{2}\cdot2-2\dfrac{5}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)
`=`\(1-\dfrac{19}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)
`=`\(-\dfrac{12}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)
`=`\(-\dfrac{3}{14}-\dfrac{10}{15}=-\dfrac{37}{42}\)
`d) `
\(\dfrac{1}{6}\cdot\dfrac{1}{11}+\dfrac{4}{11}\cdot\left(-\dfrac{1}{6}\right)+\dfrac{8}{11}\cdot\dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{6}{11}\)
`=`\(\dfrac{1}{6}\cdot\left(\dfrac{1}{11}-\dfrac{4}{11}+\dfrac{8}{11}+\dfrac{6}{11}\right)\)
`=`\(\dfrac{1}{6}\cdot\left(\dfrac{1-4+8+6}{11}\right)\)
`=`\(\dfrac{1}{6}\cdot1=\dfrac{1}{6}\)
`e)`
\(-17\cdot\left(-23\right)+\left(-53\right)\cdot17+17\cdot14+17\cdot\left(-24\right)\)
`= 17*(23-53+14-24)`
`= 17*(-40)`
`= -680`
`f)`
\(-19\cdot218+\left(-82\right)\cdot19-533\cdot19+\left(-19\right)\cdot167\)
`= 19*(-218-82-533-167)`
`= 19*(-1000)`
`= -19000`
`g)`
\(\dfrac{2}{5}+\dfrac{3}{8}-\dfrac{11}{44}+\dfrac{9}{16}\)
`=`\(\dfrac{2}{5}+\dfrac{3}{8}-\dfrac{1}{4}+\dfrac{9}{16}\)
`=`\(\dfrac{31}{40}-\dfrac{1}{4}+\dfrac{9}{16}\)
`=`\(\dfrac{21}{40}+\dfrac{9}{16}=\dfrac{87}{80}\)
`h)`
\(\dfrac{4}{10}-1\dfrac{5}{6}\cdot2+\dfrac{7}{8}-\dfrac{1}{9}\)
`=`\(\dfrac{4}{10}-\dfrac{11}{6}\cdot2+\dfrac{7}{8}-\dfrac{1}{9}\)
`=`\(\dfrac{4}{10}-\dfrac{11}{3}+\dfrac{7}{8}-\dfrac{1}{9}\)
`=`\(-\dfrac{49}{15}+\dfrac{7}{8}-\dfrac{1}{9}\)
`=`\(-\dfrac{287}{120}-\dfrac{1}{9}=-\dfrac{901}{360}\)
`i )`
\(3\cdot\dfrac{1}{5}-\dfrac{2}{8}-\dfrac{12}{36}+\dfrac{15}{9}\)
`=`\(\dfrac{3}{5}-\dfrac{1}{4}-\dfrac{1}{3}+\dfrac{15}{9}\)
`=`\(\dfrac{7}{20}-\dfrac{1}{3}+\dfrac{15}{9}\)
`=`\(\dfrac{1}{60}+\dfrac{15}{9}=-\dfrac{33}{20}\)
`k)`
\(\dfrac{6}{8}\cdot3\dfrac{1}{2}+4\dfrac{2}{3}-\dfrac{11}{55}+\dfrac{17}{51}\)
`=`\(\dfrac{3}{4}\cdot\dfrac{7}{2}+\dfrac{14}{3}-\dfrac{1}{5}+\dfrac{17}{51}\)
`=`\(\dfrac{21}{8}+\dfrac{14}{3}-\dfrac{1}{5}+\dfrac{17}{51}\)
`=`\(\dfrac{175}{24}-\dfrac{1}{5}+\dfrac{17}{51}\)
`=`\(\dfrac{851}{120}+\dfrac{17}{51}=\dfrac{297}{40}\)
`l )`
\(\dfrac{1}{3}\cdot3\dfrac{1}{2}-4\dfrac{2}{5}-\dfrac{26}{78}+\dfrac{17}{51}\)
`=`\(\dfrac{1}{3}\cdot\dfrac{7}{2}-\dfrac{22}{5}-\dfrac{1}{3}+\dfrac{17}{51}\)
`=`\(\dfrac{1}{3}\left(\dfrac{7}{2}-1\right)-\dfrac{22}{5}+\dfrac{17}{51}\)
`=`\(\dfrac{1}{3}\cdot\dfrac{5}{2}-\dfrac{22}{5}+\dfrac{17}{51}\)
`=`\(\dfrac{5}{6}-\dfrac{22}{5}+\dfrac{17}{51}\)
`=`\(-\dfrac{107}{30}+\dfrac{17}{51}=-\dfrac{97}{30}\)
P/s: Bạn tách bài ra hỏi nhé! Và ghi đề rõ ràng chứ đừng ghi ntnay, nhiều bạn nhìn vào rất khó nhìn!
`# \text {KaizulvG}`
NT
M = { 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18+ 19 + 20 }
5
21 tháng 3 2017
M = 210.
Đúng 100% luôn!
Chúc các bạn học giỏi.
Bấm đúng cho mình nha.
TN
22 tháng 7 2016
1)C= 1/5+1/10+1/20+1/40+...+1/1280
\(=\frac{1}{5}\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)
Đặt cái trong ngoặc là A ta có:\(2A=2+1+...+\frac{1}{2^7}\)
\(2A-A=\left(2+1+...+\frac{1}{2^7}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)
\(A=2-\frac{1}{2^8}\).Thay A vào ta được:\(C=\frac{1}{5}\left(2-\frac{1}{2^8}\right)=\frac{1}{5}\cdot\frac{511}{256}=\frac{511}{1280}\)
2)D= 2/1*3+2/3*5+2/5*10+2/7*9+2/9*11+2/11*18+2/13*15
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)
\(=1-\frac{1}{15}\)
\(=\frac{14}{15}\)
3)E= 4/3*7+4/7*11+4/11*15+4/15*19+4/19*23+4/23*27
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}\)
\(=\frac{8}{27}\)
4)G= 1/2+1/6+1/12+1/20+1/30+1/42+...+1/110
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
5)H= 3/1*2+3/2*3+3/3*4+3/4*5+...+3/9*10
\(=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=3\left(1-\frac{1}{10}\right)\)
\(=3\times\frac{9}{10}\)
\(=\frac{27}{10}\).Lần sau bạn đăng ít một thôi nhé
12 tháng 1 2023
a: \(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\cdot\dfrac{-7}{11}=-\dfrac{5}{11}\)
b: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\left(4+\dfrac{3}{8}\right)\)
\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{3}{2}\cdot5=\dfrac{15}{2}\)
c: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}=\dfrac{2}{15}\cdot\left(-4\right)+\dfrac{3}{15}=\dfrac{-8+3}{15}=\dfrac{-5}{15}=-\dfrac{1}{3}\)
d: \(=\dfrac{4}{9}\left(19+\dfrac{1}{3}-39-\dfrac{1}{3}\right)=\dfrac{4}{9}\cdot\left(-20\right)=-\dfrac{80}{9}\)
25 tháng 1 2022
a, \(=\dfrac{2}{9}-\dfrac{10}{10}=\dfrac{2}{9}-1=-\dfrac{7}{9}\)
b, \(=-\dfrac{12}{6}+\dfrac{2}{5}=-2+\dfrac{2}{5}=-\dfrac{8}{5}\)
c, \(=\dfrac{27}{13}-1=\dfrac{14}{13}\)
d, \(=\dfrac{12}{11}+\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+1=\dfrac{23}{11}\)
11+12-1=22
1+9+10-2-3=15
11+11+11-19=14