\(\sqrt{5-\sqrt{13+4\sqrt{3}+\sqrt{3+\sqrt{13+4\sqrt{3}}}}}\)

\(=\sqrt{5-\sqrt{13+4\sqrt{3}+\sqrt{3}+1}}\)

\(=\sqrt{5-\sqrt{14+5\sqrt{3}}}\)

\(=\dfrac{\sqrt{10-2\sqrt{14+5\sqrt{3}}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{10-\sqrt{2}\cdot\left(\sqrt{5}+3\right)}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{10-\sqrt{10}-3\sqrt{2}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{20-2\sqrt{10}-6\sqrt{2}}}{2}\)

\(\sqrt{5}-\sqrt{13}+4\sqrt{3}+\sqrt{3}+4\sqrt{3}+13\)

=27.21897397

.

a) = 1.1204
b) = 0.5714

a) = 1.1204
b) = 0.5714

a,     (\(\dfrac{9}{10}\) - \(\dfrac{15}{16}\)) \(\times\) ( \(\dfrac{5}{12}\) - \(\dfrac{11}{15}\) - \(\dfrac{7}{20}\))

=  (\(\dfrac{72}{80}\) - \(\dfrac{75}{80}\))  \(\times\) (\(\)\(\dfrac{25}{60}\) - \(\dfrac{44}{60}\)  - \(\dfrac{21}{60}\))

= - \(\dfrac{3}{80}\)  \(\times\) (- \(\dfrac{2}{3}\))

= \(\dfrac{1}{40}\) 

b, (-1)³ + (- \(\dfrac{2}{3}\))² : 2\(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)

=  -1³ +   \(\dfrac{4}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{5}{6}\)

= -1 + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{3}{8}\) + \(\dfrac{5}{6}\)

= -1 + \(\dfrac{1}{6}\) + \(\dfrac{5}{6}\)
= -1 + 1

= 0

Bài 3 là hỗn số hả em?

Bài 1: 

a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)

\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)

hay \(x=-\dfrac{1}{3}\)

Vậy: \(x=-\dfrac{1}{3}\)

b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)

\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)

hay \(x=\dfrac{50}{9}\)

Vậy: \(x=\dfrac{50}{9}\)

c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)

\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)

\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)

hay \(x=\dfrac{22}{15}\)

Vậy: \(x=\dfrac{22}{15}\)

d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)

\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)

\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)

hay \(x=\dfrac{15}{19}\)

Vậy:\(x=\dfrac{15}{19}\)

\(1,-\dfrac{4}{7}+\dfrac{2}{3}\times\dfrac{-9}{14}\)

\(=\dfrac{-4}{7}+\dfrac{-18}{42}\)

\(=\dfrac{-4\times6}{7\times6}+\dfrac{-18}{42}\)

\(=\dfrac{-20}{42}+\dfrac{-18}{42}\)

\(=-\dfrac{38}{42}\)

\(=-\dfrac{19}{21}\)

\(2,\dfrac{17}{13}-\left(\dfrac{4}{13}-11\right)\)

\(=\dfrac{17}{13}-\dfrac{4}{13}+11\)

\(=\dfrac{13}{13}+11\)

\(=1+11\)

\(=12\)

\(3,8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)

\(=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)

\(=\dfrac{58}{7}-\dfrac{31}{9}-\dfrac{30}{7}\)

\(=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}\)

\(=\dfrac{28}{7}-\dfrac{31}{9}\)

\(=\dfrac{28\times9}{7\times9}-\dfrac{31\times7}{9\times7}\)

\(=\dfrac{252}{63}-\dfrac{217}{63}\)

\(=\dfrac{35}{63}\)

\(=\dfrac{5}{9}\)

\(5,\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)

\(=\left(\dfrac{2}{3}-\dfrac{3}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)

\(=\left(\dfrac{2\times2}{3\times2}-\dfrac{3\times3}{2\times3}\right):\dfrac{4}{3}+\dfrac{1}{2}\)

\(=\left(\dfrac{4}{6}-\dfrac{9}{6}\right):\dfrac{4}{3}+\dfrac{1}{2}\)

\(=\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}\)

\(=\dfrac{-5}{6}\times\dfrac{3}{4}+\dfrac{1}{2}\)

\(=\dfrac{-15}{24}+\dfrac{1}{2}\)

\(=\dfrac{-15}{24}+\dfrac{1\times12}{2\times12}\)

\(=\dfrac{-15}{24}+\dfrac{12}{24}\)

\(=\dfrac{-3}{24}\)

\(=-\dfrac{1}{8}\)

\(6,\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}\)

\(=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}\)

\(=\dfrac{-13}{13}+\dfrac{5}{5}-\dfrac{3}{7}\)

\(=-1+1-\dfrac{3}{7}\)

\(=-\dfrac{3}{7}\)

\(7,\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}:\dfrac{7}{10}+\dfrac{6}{5}\)

\(=\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}\times\dfrac{10}{7}+\dfrac{6}{5}\)

\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+1\right)\)

\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{1\times7}{1\times7}\right)\)

\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{7}{7}\right)\)

\(=\dfrac{6}{5}\times\dfrac{20}{7}\)

\(=\dfrac{120}{35}\)

\(=\dfrac{24}{7}\)

\(-\frac{3}{7}.\frac{5}{9}+\frac{4}{9}.\frac{-3}{7}+2.\frac{3}{7}\)\(=\frac{3}{7}\left(-\frac{5}{9}-\frac{4}{9}+2\right)=\frac{3}{7}\left(-1+2\right)=\frac{3}{7}\)

so sánh: \(\frac{21}{-42}=\frac{1}{-2}=\frac{2}{-4}\); vì \(\frac{2}{-4}>\frac{3}{-4}\)nên \(\frac{21}{-42}>\frac{3}{-4}\)

tìm x: a) \(x-\frac{2}{5}=\frac{1}{10}\)

             \(\Rightarrow x=\frac{1}{10}+\frac{2}{5}=\frac{1}{10}+\frac{4}{10}=\frac{5}{10}=\frac{1}{2}\)

b) \(\frac{4}{6}+\frac{5}{4}x=\frac{7}{6}\)

\(\Rightarrow\frac{5}{4}x=\frac{3}{6}=\frac{1}{2}\)

\(\Rightarrow x=\frac{2}{5}\)

Bài 4 là:

\(\dfrac{1}{2.5}\) + \(\dfrac{1}{5.8}\) +....+\(\dfrac{1}{92.95}\) + \(\dfrac{1}{95.98}\)

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