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16 tháng 1 2019

a)  5 8 − 4 9 = 45 72 − 32 72 = 13 72

b)  1 2 : 7 8 = 1 2 × 8 7 = 4 7

17 tháng 4 2022

a) 4 + 5/7 = .... 28/7 + 5/7=.33/7..............

    5/9 x 6/7 = ..10/21....

    3 - 7/5 =  15/5 - 7/5...8/5........

    3/5 x 4/8 = ..3/10....

b) 2 - 1/4 = .8/4 - 1/4...7/4...

    4 : 5/9 = ..4 x 9/5 = 36/5....

    2/9 x 3/5 = .....2/15.....

    3/8 : 4 = ......3/8 x 1/4 = 3/2.....

17 tháng 4 2022

Thanks kiuuuuuuu

a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)

\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)

\(=\dfrac{-1621}{126}\)

b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)

\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)

\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)

\(=-\dfrac{49}{20}\)

a: \(=\dfrac{6}{7}\cdot\dfrac{-3}{5}=\dfrac{-18}{35}\)

b: \(=\dfrac{2}{5}\cdot\dfrac{-15}{8}=\dfrac{-30}{40}=-\dfrac{3}{4}\)

c: \(=\dfrac{2}{4}\cdot\dfrac{7}{3}=\dfrac{1}{2}\cdot\dfrac{7}{3}=\dfrac{7}{6}\)

d: \(=\dfrac{8}{3}\cdot\dfrac{16}{4}=\dfrac{128}{12}=\dfrac{32}{3}\)

24 tháng 1 2022

a)-18/35

b)-3/4

c)7/6

d)32/3

e)5/21

sai thì xin lỗi nhé bn

21 tháng 5 2023

a)

\(M=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{4+4\sqrt{5}+5}-\sqrt{4-4\sqrt{5}+5}\)

\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)

\(=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|\)

\(=2+\sqrt{5}-\left(\sqrt{5}-2\right)\) (vì \(2+2\sqrt{5}>0;2-\sqrt{5}< 0\) )

\(=2+\sqrt{5}-\sqrt{5}+2\\ =4\)

b)

\(N=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)

\(=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)

\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|\)

\(=\sqrt{7}-1-\left(\sqrt{7}+1\right)\) (vì \(\sqrt{7}-1>0;\sqrt{7}+1>0\) )

\(=\sqrt{7}-1-\sqrt{7}-1\\ =-2\)

\(\dfrac{3}{7}\times\dfrac{7}{9}\times\dfrac{1}{2}\)

\(=\dfrac{3\times7\times1}{7\times9\times2}\)

\(=\dfrac{21}{126}\)

\(=\dfrac{1}{6}\)

\(\dfrac{5}{8}\times4\times\dfrac{1}{2}\\ =\dfrac{5}{8}\times\dfrac{4}{1}\times\dfrac{1}{2}\\ =\dfrac{5\times4\times1}{8\times1\times2}\\ =\dfrac{20}{16}\\ =\dfrac{5}{4}\)

\(4\times\dfrac{1}{24}\times3\\ =\dfrac{4}{1}\times\dfrac{1}{24}\times\dfrac{3}{1}\\ =\dfrac{4\times1\times3}{1\times24\times1}\\ =\dfrac{12}{24}\\ =\dfrac{1}{2}\)