Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a; \(\dfrac{3}{11}\) + \(\dfrac{5}{-9}\) + \(\dfrac{4}{11}\) - \(\dfrac{4}{9}\) + \(\dfrac{3}{17}\) + \(\dfrac{15}{11}\)
= (\(\dfrac{3}{11}\) + \(\dfrac{4}{11}\) + \(\dfrac{15}{11}\)) - (\(\dfrac{5}{9}\) + \(\dfrac{4}{9}\)) + \(\dfrac{3}{17}\)
= 2 - 1 + \(\dfrac{3}{17}\)
= 1 + \(\dfrac{3}{17}\)
= \(\dfrac{20}{17}\)
c; N = \(\dfrac{\dfrac{5}{7}-\dfrac{5}{9}-\dfrac{5}{11}}{\dfrac{15}{7}+\dfrac{15}{9}+\dfrac{15}{11}}\)
Phải là - \(\dfrac{5}{7}\) chỗ tử số mới đúng em nhé!
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
`@` `\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}`
a,
A=1−3−5−7−9−...−97−99a)A=1−3−5−7−9−...−97−99
=1−(3+5+7+...+99)=1−(3+5+7+...+99)
=1−(99+3).[(99−3):2+1]2=1−(99+3).[(99−3):2+1]2
=1−2499=−2498=1−2499=−2498
b)B=1+3−5−7+9+...+97−99b)B=1+3−5−7+9+...+97−99
=(−8)+(−8)+(−8)+...+(−8)+97−99=(−8)+(−8)+(−8)+...+(−8)+97−99
=(−8).12+(−2)=−98=(−8).12+(−2)=−98
c)C=1−3−5+7+9−11−13+15+...+97−99c)C=1−3−5+7+9−11−13+15+...+97−99
=0+0+0+0+0+...+0−99=0+0+0+0+0+...+0−99
=−99