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Dấu " / " là phân số nhé
a) 5/-4 . 16/25 + -5/4 . 9/25
= -5/4 . 16/25 + -5/4 . 9/25
= -5/4 . ( 16/25 + 9/25 )
= -5/4 . 1
= -5/4
b) 4 11/23 - 9/14 + 2 12/23 - 5/4
= 103/23 - 9/14 + 58/23 - 5/4
= 103/23 + 58/23 - 9/14 - 5/4
= 7 - 9/14 - 5/4
= 143/28
c) 2 13/27 - 7/15 + 3 14/27 - 8/15
= 67/27 - 7/15 + 95/27 - 8/15
= 67/27 + 95/27 - 7/15 - 8/15
= 6 - 7/15 - 8/15
= 5
1) âm năm phần 12
2) âm mười bảy phần 9
3) -1
Đây là đáp án còn làm bài từ làm nhé
\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}\)
\(A=\dfrac{2}{4}-\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\)
\(D=\dfrac{2}{3\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{11}{8\cdot19}+\dfrac{13}{19\cdot32}+\dfrac{25}{32\cdot57}+\dfrac{30}{57\cdot87}\)
Áp dụng công thức tổng quát \(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
Ta có:
\(D=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{57}+\dfrac{1}{57}-\dfrac{1}{87}\\ D=\dfrac{1}{3}-\dfrac{1}{87}\\ D=\dfrac{28}{87}\)
2:
a: x=2,4-0,4=2
b: =>2x=-1,5+0,8=-0,7
=>x=-0,35
c: =>x-16=-15
=>x=1
\(\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{50-\dfrac{2}{11}+\dfrac{8}{12}-\dfrac{8}{15}}=\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{2\left(25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}\right)}\)
\(=\dfrac{1}{2}\)
\(\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{50-\dfrac{2}{11}+\dfrac{8}{13}-\dfrac{8}{15}}=\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{2.\left(25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}\right)}\)
A = \(\dfrac{1}{4}.\dfrac{7}{3}.12\)
= \(\dfrac{1.7.12}{4.3}\)
= \(7\)
@Nguyễn Thành Đăng
B = \(\dfrac{3}{8}.56.\dfrac{25}{7}.\left(-4\right)\)
= \(-\dfrac{3.56.25.4}{8.7}\)
= -3.100
= -300
@Nguyễn Thành Đăng
A,
\(\left(7\dfrac{4}{9}+3\dfrac{7}{11}\right)-3\dfrac{4}{9}=7\dfrac{4}{9}+3\dfrac{7}{11}-3\dfrac{4}{9}\)
\(=7\dfrac{4}{9}-3\dfrac{4}{9}+3\dfrac{7}{11}=4+3\dfrac{7}{11}=7\dfrac{7}{11}\)
B,
\(5\dfrac{2}{7}.\dfrac{8}{11}+5\dfrac{2}{7}.\dfrac{5}{11}-5\dfrac{2}{7}.\dfrac{2}{11}=5\dfrac{2}{7}.\left(\dfrac{8}{11}+\dfrac{5}{11}-\dfrac{2}{11}\right)\)
\(=5\dfrac{2}{7}.1=5\dfrac{2}{7}\)
Giải:
h) \(\left(-\dfrac{1}{2}\right)^2.13\dfrac{9}{11}-25\%.6\dfrac{2}{11}\)
\(=\dfrac{1}{4}.\dfrac{152}{11}-\dfrac{1}{4}.\dfrac{68}{11}\)
\(=\dfrac{1}{4}\left(\dfrac{152}{11}-\dfrac{68}{11}\right)\)
\(=\dfrac{1}{4}.\dfrac{84}{11}=\dfrac{21}{11}\)
Vậy ...
i) \(57\%+1\dfrac{2}{15}:\dfrac{68}{25}-2,82\)
\(=\dfrac{57}{100}+\dfrac{17}{15}:\dfrac{68}{25}-\dfrac{141}{50}\)
\(=\dfrac{57}{100}+\dfrac{2}{15}-\dfrac{141}{50}\)
\(=\dfrac{-127}{60}\)
Vậy ...
h. \(\left(-\dfrac{1}{2}\right)^2\times13\dfrac{9}{11}-25\%\times6\dfrac{2}{11}\)
= \(\dfrac{1}{4}\times\dfrac{150}{11}-\dfrac{1}{4}\times34\)
= \(\dfrac{75}{22}-\dfrac{17}{2}\)
= \(\dfrac{75}{22}-\dfrac{187}{22}\)
= \(-\dfrac{56}{11}\)
i. \(57\%+1\dfrac{2}{15}\div\dfrac{68}{25}-2,82\)
= \(\dfrac{57}{100}+\dfrac{17}{15}\div\dfrac{68}{25}-\dfrac{282}{100}\)
= \(\dfrac{57}{100}+\dfrac{17}{15}\times\dfrac{25}{68}-\dfrac{282}{100}\)
= \(\dfrac{57}{100}+\dfrac{5}{3}-\dfrac{282}{100}\)
=\(\dfrac{5}{3}-\left(\dfrac{57}{100}-\dfrac{282}{100}\right)\)
= \(\dfrac{5}{3}-\left(-\dfrac{215}{100}\right)\)
= \(\dfrac{5}{3}-\left(-\dfrac{43}{20}\right)\)
= \(\dfrac{5}{3}+\dfrac{43}{30}\)
= \(\dfrac{50}{30}+\dfrac{43}{30}\)
= \(\dfrac{31}{10}\)