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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
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
\(D=\dfrac{2}{3\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{11}{8\cdot19}+\dfrac{13}{19\cdot32}+\dfrac{25}{32\cdot57}+\dfrac{30}{57\cdot87}\)
Áp dụng công thức tổng quát \(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
Ta có:
\(D=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{57}+\dfrac{1}{57}-\dfrac{1}{87}\\ D=\dfrac{1}{3}-\dfrac{1}{87}\\ D=\dfrac{28}{87}\)
a)\(\dfrac{-5}{9}+\dfrac{5}{9}:\left(1\dfrac{2}{3}-2\dfrac{1}{12}\right)\)
\(=\dfrac{-5}{9}+\dfrac{5}{9}:\left(\dfrac{20}{12}-\dfrac{25}{12}\right)\)
\(=\dfrac{-5}{9}+\dfrac{5}{9}\cdot\dfrac{-12}{5}\)
\(=\dfrac{-5}{9}+\dfrac{-12}{9}\)
\(=\dfrac{-17}{9}\)
b) \(\dfrac{-7}{25}\cdot\dfrac{11}{13}+\dfrac{-7}{25}\cdot\dfrac{2}{13}-\dfrac{18}{25}\)
\(=\dfrac{-7}{25}\cdot\left(\dfrac{11}{13}+\dfrac{2}{13}\right)-\dfrac{18}{25}\)
\(=\dfrac{-7}{25}\cdot1-\dfrac{18}{25}\)
\(=\dfrac{-25}{25}\)
\(=-1\)
c) \(\left(\dfrac{5}{7}\cdot0,6-5:3\dfrac{1}{2}\right)\cdot\left(40\%-1,4\right)\cdot\left(-2\right)^3\)
\(=\left(\dfrac{3}{7}-\dfrac{10}{7}\right)\cdot\left(0,4-1,4\right)\cdot\left(-8\right)\)
\(=\left(-1\right)\cdot\left(-1\right)\cdot\left(-8\right)\)
\(=-8\)
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,
\(5\dfrac{2}{7}.\dfrac{8}{11}+5\dfrac{2}{7}.\dfrac{5}{11}-5\dfrac{2}{7}.\dfrac{2}{11}=5\dfrac{2}{7}.\left(\dfrac{8}{11}+\dfrac{5}{11}-\dfrac{2}{11}\right)\)
\(=5\dfrac{2}{7}.1=5\dfrac{2}{7}\)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
a) \(\dfrac{11}{21}+\dfrac{-4}{7}=\dfrac{11}{21}+\dfrac{-12}{21}=\dfrac{-1}{21}\)
b) \(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}=\dfrac{1}{3}+\dfrac{14}{25}-\dfrac{4}{3}+\dfrac{2}{7}+\dfrac{11}{25}\)
\(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)+\dfrac{2}{7}=-1+1+\dfrac{2}{7}=\dfrac{2}{7}\)
c) \(\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{3}{14}=\dfrac{28}{42}+\dfrac{30}{42}-\dfrac{9}{42}=\dfrac{49}{42}=\dfrac{7}{6}\)
d) \(\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{9}{45}=\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{1}{5}=\dfrac{14}{35}-\dfrac{15}{35}+\dfrac{7}{35}=\dfrac{6}{35}\)
e) \(\dfrac{21}{47}+\dfrac{9}{45}+\dfrac{26}{47}+\dfrac{45}{5}=\dfrac{21}{47}+\dfrac{1}{5}+\dfrac{26}{47}+\dfrac{45}{5}=\left(\dfrac{21}{47}+\dfrac{26}{47}\right)+\left(\dfrac{1}{5}+\dfrac{45}{5}\right)\)
\(=1+\dfrac{46}{5}=\dfrac{51}{5}\)
f) \(\dfrac{15}{12}-\dfrac{18}{13}+\dfrac{5}{13}-\dfrac{3}{12}=\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(-\dfrac{18}{13}+\dfrac{5}{13}\right)=1+\left(-1\right)=0\)
g) \(\dfrac{-8}{18}-\dfrac{15}{27}=\dfrac{-4}{9}-\dfrac{5}{9}=\dfrac{-9}{9}=-1\)
h)\(\dfrac{3}{7}+\dfrac{-5}{2}-\dfrac{3}{5}=\dfrac{30}{70}+\dfrac{-175}{70}-\dfrac{42}{70}=\dfrac{-187}{70}\)
i) \(\left(\dfrac{11}{12}:\dfrac{33}{16}\right).\dfrac{3}{5}=\dfrac{11}{12}.\dfrac{16}{33}.\dfrac{3}{5}=\dfrac{11.16.3}{12.33.5}=\dfrac{4}{15}\)
Đây này má Ran mori
a) \(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
\(=5+\dfrac{1}{7}-3-\dfrac{3}{11}-2-\dfrac{1}{7}-1-\dfrac{8}{11}\)
\(=\left(5-3-2-1\right)+\left(\dfrac{1}{7}-\dfrac{3}{11}-\dfrac{1}{7}-\dfrac{8}{11}\right)\)
\(=-1+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(=-1+0-1=-2\)
a)\(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
= \(\left(5+\dfrac{1}{7}-3+\dfrac{3}{11}\right)-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(5-\dfrac{1}{7}+3-\dfrac{3}{11}-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(\left(5-3-2-1\right)+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{8}{11}-\dfrac{3}{11}\)
= \(-1+2+\dfrac{5}{11}\)
= \(1+\dfrac{5}{11}=\dfrac{1}{1}+\dfrac{5}{11}=\dfrac{11}{11}+\dfrac{5}{11}=\dfrac{16}{11}\)
Vậy :câu a) = \(\dfrac{16}{11}\)
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. \(57\%+1\dfrac{2}{15}\div\dfrac{68}{25}-2,82\)
= \(\dfrac{57}{100}+\dfrac{17}{15}\div\dfrac{68}{25}-\dfrac{282}{100}\)
= \(\dfrac{57}{100}+\dfrac{17}{15}\times\dfrac{25}{68}-\dfrac{282}{100}\)
= \(\dfrac{57}{100}+\dfrac{5}{3}-\dfrac{282}{100}\)
=\(\dfrac{5}{3}-\left(\dfrac{57}{100}-\dfrac{282}{100}\right)\)
= \(\dfrac{5}{3}-\left(-\dfrac{215}{100}\right)\)
= \(\dfrac{5}{3}-\left(-\dfrac{43}{20}\right)\)
= \(\dfrac{5}{3}+\dfrac{43}{30}\)
= \(\dfrac{50}{30}+\dfrac{43}{30}\)
= \(\dfrac{31}{10}\)