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19 tháng 12 2021

\(=\dfrac{1}{2}\cdot10-\dfrac{1}{4}+1=6-\dfrac{1}{4}=\dfrac{23}{4}\)

23 tháng 8 2021

a)\(\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{3}{5}=\dfrac{15}{20}-\dfrac{50}{20}-\dfrac{12}{20}=-\dfrac{47}{20}\)

b) \(\sqrt{7^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}=7+\sqrt{\dfrac{1}{16}}=7+\dfrac{1}{4}=\dfrac{29}{4}\)

c) \(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}=\dfrac{1}{2}.10-\sqrt{\dfrac{1}{16}+1}=5-\sqrt{\dfrac{17}{16}}\)

25 tháng 10 2022

a: \(=7\cdot\dfrac{6}{7}-5+\dfrac{3\sqrt{2}}{2}=1+\dfrac{3}{2}\sqrt{2}\)

b: \(=-\dfrac{8}{7}-\dfrac{3}{5}\cdot\dfrac{5}{8}+\dfrac{1}{2}=\dfrac{-16+7}{14}-\dfrac{3}{8}=\dfrac{-9}{14}-\dfrac{3}{8}\)

\(=\dfrac{-72-42}{112}=\dfrac{-114}{112}=-\dfrac{57}{56}\)

c: \(=20\sqrt{5}-\dfrac{1}{4}\cdot\dfrac{4}{3}+\dfrac{3}{2}=20\sqrt{5}+\dfrac{3}{2}-\dfrac{1}{3}=20\sqrt{5}+\dfrac{7}{6}\)

 

20 tháng 12 2022

\(0,5\sqrt{100}-\dfrac{1}{4}\sqrt{16}+\left(-\dfrac{2}{3}\right)^2\\=\dfrac{1}{2}.10-\dfrac{1}{4}.4+\dfrac{4}{9}\\ =5-1+\dfrac{4}{9}\\ =4+\dfrac{4}{9}\\ =\dfrac{40}{9} \)

20 tháng 12 2022

`(-2/3)^2` Ạ

3 tháng 12 2018

\(x=\left(1-\dfrac{1}{\sqrt{4}}\right).\left(1-\dfrac{1}{\sqrt{16}}\right).\left(1-\dfrac{1}{\sqrt{36}}\right).\left(1-\dfrac{1}{\sqrt{64}}\right).\left(1-\dfrac{1}{\sqrt{100}}\right)\)

\(x=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{6}\right).\left(1-\dfrac{1}{8}\right).\left(1-\dfrac{1}{10}\right)\)

\(x=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.\dfrac{7}{8}.\dfrac{9}{10}\)

\(x=\dfrac{63}{256}\)

\(y=\sqrt{20+0,25}\)

\(y=\sqrt{20,25}\)

\(y=4,5\)

Do 4,5 > \(\dfrac{63}{256}\)

=> x<y

4 tháng 12 2018

cho mình hỏi tại sao 4,5 > \(\dfrac{63}{256}\)

25 tháng 10 2022

a: \(=2\cdot\dfrac{5}{4}-3\cdot\dfrac{7}{6}+4\cdot\dfrac{9}{8}=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{9}{2}=\dfrac{7}{2}\)

b: \(=18-16\cdot\dfrac{1}{2}+\dfrac{1}{16}\cdot\dfrac{3}{4}\)

=10+3/64

=643/64

c: \(=\dfrac{2}{3}\cdot\dfrac{9}{4}-\dfrac{3}{4}\cdot\dfrac{8}{3}+\dfrac{7}{5}\cdot\dfrac{5}{14}=\dfrac{3}{2}-2+\dfrac{1}{2}=2-2=0\)

17 tháng 12 2021

\(\sqrt{16}.\left(\dfrac{1}{2}\right)^3+3.\left(\dfrac{1}{2}\right)^2+2.\sqrt{\dfrac{1}{4}}\)
\(\text{​​}4.\left(-\dfrac{1}{8}\right)+3.\left(\dfrac{1}{4}\right)+2.\left(\dfrac{1}{2}\right)\)
\(-\dfrac{1}{2}+\dfrac{3}{4}+1=\dfrac{5}{4}\)

\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)

\(=\dfrac{4}{7}+\dfrac{1}{8}-\dfrac{4}{7}-\dfrac{7}{8}\)

\(=\dfrac{1}{8}-\dfrac{7}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}\)

\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0,5\left(-2\dfrac{3}{5}\right)\)

\(=\left|\dfrac{5-6}{10}\right|\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)

\(=\dfrac{1}{10}\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)

\(=\dfrac{3}{10}-\dfrac{13}{10}=-\dfrac{10}{10}=-1\)