1+1+3+9+17
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PT
4 tháng 4 2016
\((\frac{1}{9}-\frac{9}{17})+\frac{3}{6}-(\frac{-8}{17}-\frac{1}{2})+\frac{5}{9}=\)
=\(\frac{1}{9}-\frac{9}{17}+\frac{1}{2}+\frac{8}{17}+\frac{1}{2}+\frac{5}{9}\)
= \((\frac{1}{9}+\frac{5}{9})+(\frac{-9}{17}+\frac{8}{17})+(\frac{1}{2}+\frac{1}{2})\)
=\(\frac{2}{3}-\frac{1}{17}+1\)
=\(\frac{34}{51}-\frac{3}{51}+\frac{51}{51}\)
=\(\frac{82}{51}\)
NT
0
17 tháng 8 2023
1:
a: =23/27-11/17+4/27+28/17
=23/27+4/27+28/17-11/17
=1+1=2
b: \(=\dfrac{2}{3}\cdot\left(\dfrac{7}{9}+\dfrac{2}{9}\right)-\dfrac{2}{9}\)
=2/3-2/9
=6/9-2/9
=4/9
c: \(=\dfrac{11}{5}\cdot\dfrac{7}{3}-\dfrac{1}{3}\cdot\dfrac{11}{5}\)
=11/5(7/3-1/3)
=11/5*2
=22/5
d: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2024}{2023}=\dfrac{2024}{2}=1012\)
e: \(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2022}{2023}=\dfrac{1}{2023}\)
16 tháng 2 2018
bài 1.
A) \(\dfrac{-5}{9}+\dfrac{3}{5}-\dfrac{3}{9}+\dfrac{-2}{5}=\left(\dfrac{-5}{9}-\dfrac{3}{9}\right)+\left(\dfrac{3}{5}+\dfrac{-2}{5}\right)=\dfrac{-8}{9}+\dfrac{1}{5}=-\dfrac{31}{45}\)
B)\(-\dfrac{5}{13}+\left(\dfrac{3}{5}+\dfrac{3}{1}-\dfrac{4}{10}\right)=-\dfrac{5}{13}+3+\left(\dfrac{3}{5}-\dfrac{4}{10}\right)=-\dfrac{5+3.13}{13}+\dfrac{1}{5}=-\dfrac{207}{65}\)C)\(\dfrac{5}{17}-\dfrac{9}{15}-\dfrac{2}{-17}+\dfrac{-3}{15}=\left(\dfrac{5}{17}+\dfrac{2}{17}\right)+\left(\dfrac{-9}{15}+\dfrac{-3}{15}\right)=\dfrac{7}{17}-\dfrac{12}{15}=-\dfrac{33}{85}\)D)\(\left(\dfrac{1}{9}-\dfrac{9}{17}\right)+\dfrac{3}{6}-\left(\dfrac{12}{17}-\dfrac{1}{2}\right)+\dfrac{-1}{9}=\left(\dfrac{1}{9}+\dfrac{-1}{9}\right)+\left(\dfrac{3}{6}+\dfrac{1}{2}\right)-\dfrac{12}{17}=1-\dfrac{12}{17}=\dfrac{5}{17}\)
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}`
12 tháng 10 2021
Câu 2:
\(\dfrac{7}{9}\cdot\dfrac{3}{35}=\dfrac{1}{5}\cdot\dfrac{1}{3}=\dfrac{1}{15}\)
\(\dfrac{9}{22}\cdot55=\dfrac{9\cdot55}{22}=\dfrac{9\cdot5}{2}=\dfrac{45}{2}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
4 tháng 8 2023
A = \(\dfrac{4}{3}\) + \(\dfrac{17}{3}\) + \(\dfrac{17}{9}\) + \(\dfrac{19}{13}\) + \(\dfrac{1}{9}\) + \(\dfrac{14}{6}\)
A = (\(\dfrac{4}{3}\) + \(\dfrac{17}{3}\)) + (\(\dfrac{17}{9}\) + \(\dfrac{1}{9}\)) + (\(\dfrac{19}{13}\) + \(\dfrac{14}{6}\))
A = \(\dfrac{21}{3}\) + \(\dfrac{18}{9}\) + \(\dfrac{148}{39}\)
A = 7 + 2 + \(\dfrac{148}{39}\)
A = 9 + \(\dfrac{148}{39}\)
A = \(\dfrac{39\times9}{39}\) + \(\dfrac{148}{39}\)
A = \(\dfrac{499}{39}\)
H9
HT.Phong (9A5)
CTVHS
17 tháng 8 2023
Ta có:
Tập hợp A:
\(A=\left\{1;5;9;13;17;21;25\right\}\)
Tập hợp B:
\(B=\left\{0;1;3;5;10;13\right\}\)
Mà: \(A\cap B\)
\(\Rightarrow A\cap B=\left\{1;5;13\right\}\)
⇒ Chọn B
à mình nhớ rôì
=31
HÉT