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a) \(\dfrac{5}{9}:\left(\dfrac{13}{7}+\dfrac{13}{9}\right)-\dfrac{5}{3}\)(chỗ này mk lười chép lại đề)
=\(\dfrac{5}{9}:\dfrac{208}{63}-\dfrac{5}{3}\)
=\(\dfrac{5}{9}.\dfrac{63}{208}-\dfrac{5}{3}\)
=\(\dfrac{5.63}{9.208}-\dfrac{5}{3}\)
=\(\dfrac{5.7}{1.208}-\dfrac{5}{3}\)
=\(\dfrac{36}{208}-\dfrac{5}{3}\)
=\(\dfrac{108}{624}-\dfrac{1040}{624}\)
=\(\dfrac{-932}{624}\)
=\(\dfrac{233}{156}\)
còn câu b mk chưa học nên mk chịu
Giải:
5/9:13/7+5/9:13/9 -1 2/3
=5/9.7/13+5/9.9/13-5/3
=5/9.(7/13+9/13)-5/3
=5/9.16/13-5/3
=80/117-5/3
=-115/117
4 2/5 : 0,5% -1 3/7 .14% +(-0,5)
=22/5:1/200-10/7.7/50 +(-1/2)
=880-1/5-1/2
=8793/10
a)A=\(\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{-3}{5}+\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)+\left(\dfrac{-7}{9}+\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(\dfrac{-11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)
A=0+0+0+...+0+\(\dfrac{13}{15}\)
A=\(\dfrac{13}{15}\)
b) Ta có: \(-\dfrac{1}{9\cdot10}-\dfrac{1}{8\cdot9}-\dfrac{1}{7\cdot8}-...-\dfrac{1}{2\cdot3}-\dfrac{1}{1\cdot2}\)
\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=-\left(1-\dfrac{1}{10}\right)=\dfrac{-9}{10}\)
a: \(=\dfrac{37}{4}+\dfrac{117}{16}+\dfrac{1}{4}=\dfrac{19}{2}+\dfrac{117}{16}=\dfrac{269}{16}\)
b: \(=1+\left(\dfrac{9}{10}+\dfrac{8}{10}\right):\dfrac{19}{6}=1+\dfrac{17}{10}\cdot\dfrac{6}{19}=\dfrac{146}{95}\)
c: \(=\dfrac{1}{4}-\dfrac{6}{4}+\dfrac{6}{5}=\dfrac{-5}{4}+\dfrac{6}{5}=\dfrac{-1}{20}\)
\(\dfrac{2}{3}+\dfrac{1}{5}.\dfrac{10}{7}=\dfrac{2}{3}+\dfrac{10}{35}=\dfrac{70}{105}+\dfrac{30}{105}=\dfrac{100}{105}=\dfrac{50}{21}\)
a) Ta có: \(\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\)
\(=\dfrac{2}{3}+\dfrac{2}{7}\)
\(=\dfrac{14}{21}+\dfrac{6}{21}\)
\(=\dfrac{20}{21}\)
1) âm năm phần 12
2) âm mười bảy phần 9
3) -1
Đây là đáp án còn làm bài từ làm nhé
bài 3:
\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{2021\cdot2022}\)
\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2021\cdot2022}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)
\(=9\cdot\dfrac{2021}{2022}=\dfrac{6063}{674}\)
Bài 1:
a: \(\left(\dfrac{1}{2}+\dfrac{16}{30}\right)-\left(1+\dfrac{1}{30}\right)\)
\(=\dfrac{15+16}{30}-1-\dfrac{1}{30}\)
\(=\dfrac{30}{30}-1\)
=1-1
=0
b: \(\dfrac{-5}{11}\cdot\dfrac{4}{13}+\dfrac{-5}{11}\cdot\dfrac{9}{13}+3\dfrac{5}{11}\)
\(=-\dfrac{5}{11}\left(\dfrac{4}{13}+\dfrac{9}{13}\right)+3+\dfrac{5}{11}\)
\(=-\dfrac{5}{11}+3+\dfrac{5}{11}\)
=3
c: \(3^2-12\left(\dfrac{3}{4}-\dfrac{2}{3}\right)\)
\(=9-12\cdot\dfrac{9-8}{12}\)
=9-1
=8
`[-5]/13 . 2/11+[-5]/11 . 9/13+1`
`=[-5]/13 . 2/11+[-5]/13 . 9/11+1`
`=[-5]/13 . (2/11+9/11)+1`
`=[-5]/13 . 11/11+1`
`=[-5]/13+1=[-5]/13+13/13=8/13`
\(\dfrac{-5}{13}.\dfrac{2}{11}+\dfrac{-5}{13}.\dfrac{9}{13}+1\)
\(=\dfrac{-5}{13}+\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1\)
\(=\dfrac{-5}{13}.1+1\)
\(=\dfrac{-5}{13}+1\)
\(=\dfrac{8}{13}\)