a: \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}-\dfrac{8}{13}\right)+\left(\dfrac{-18}{35}-\dfrac{17}{35}\right)\)

=1-1-1

=-1

b: \(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}=\dfrac{-3}{8}-\dfrac{5}{8}=-1\)

c: \(=\dfrac{4}{4}\cdot\dfrac{5}{15}\cdot\dfrac{11}{11}=\dfrac{1}{3}\)

a) \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(-\dfrac{5}{13}+-\dfrac{8}{13}\right)+\left(-\dfrac{18}{35}+-\dfrac{17}{35}\right)=1+-1+-1=-1\)

b) \(=-\dfrac{3}{8}\cdot\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}-\dfrac{10}{16}=-1\)

c) \(=\left(-\dfrac{4}{11}\cdot-\dfrac{11}{4}\right)\cdot\dfrac{5}{15}=1\cdot\dfrac{1}{3}=\dfrac{1}{3}\)

a)\(=\left(-\dfrac{5}{13}+\dfrac{-8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)+\left(\dfrac{3}{14}+\dfrac{14}{17}\right)=-1-1+1=-1\)

b)\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}.1-\dfrac{10}{16}=-\dfrac{6}{16}-\dfrac{10}{16}=-\dfrac{16}{16}=-1\)

c)\(\dfrac{-4.5.11}{11.5.3.-4}=\dfrac{1}{3}\)

\(=\sqrt{\dfrac{4}{25}+\dfrac{4}{5}-\dfrac{9}{5}\cdot\dfrac{1}{9}+\dfrac{3}{4}}=\sqrt{\dfrac{4}{25}+\dfrac{4}{5}-\dfrac{1}{5}+\dfrac{3}{4}}\)

\(=\sqrt{\dfrac{4}{25}+\dfrac{3}{5}+\dfrac{3}{4}}\)

\(=\sqrt{\dfrac{16+60+75}{100}}=\dfrac{\sqrt{151}}{10}\)

bạn viết đúng đề chứ ạ

\(\sqrt{\dfrac{4}{25}+\left|-\dfrac{4}{5}\right|-\dfrac{9}{5}.\left(\dfrac{-1}{3}\right)^2+0,75}\)

\(=\sqrt{\dfrac{4}{25}+\dfrac{4}{5}-\dfrac{9}{5}.\dfrac{1}{9}+\dfrac{3}{4}}=\sqrt{\dfrac{4}{25}+\dfrac{20}{25}-\dfrac{9}{45}+\dfrac{3}{4}}=\sqrt{\dfrac{24}{25}-\dfrac{9}{45}+\dfrac{3}{4}}\)

\(=\sqrt{\dfrac{19}{25}+\dfrac{3}{4}}=\sqrt{\dfrac{76}{100}+\dfrac{75}{100}}=\sqrt{\dfrac{151}{100}}=\dfrac{\sqrt{151}}{10}\)

a: \(=\dfrac{13\left(3-18\right)}{40\left(15-2\right)}=\dfrac{13}{15-2}\cdot\dfrac{-15}{40}=\dfrac{-3}{8}\)

b: \(=\dfrac{18\left(34-124\right)}{36\left(-17-13\right)}=\dfrac{1}{2}\cdot\dfrac{-90}{-30}=\dfrac{3}{2}\)

c: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{\dfrac{-1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)

\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}\)

\(=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)

\(a.\)

\(-\dfrac{2}{3}\cdot\dfrac{?}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{?}{4}=\dfrac{1}{2}:-\dfrac{2}{3}=\dfrac{1}{2}\cdot-\dfrac{3}{2}=-\dfrac{3}{4}\)

\(\Leftrightarrow?=-3\)

\(b.\)

\(\dfrac{?}{3}\cdot\dfrac{5}{8}=-\dfrac{5}{12}\)

\(\Leftrightarrow\dfrac{?}{3}=\dfrac{-5}{12}:\dfrac{5}{8}=\dfrac{-5}{12}\cdot\dfrac{8}{5}=-\dfrac{2}{3}\)

\(\Leftrightarrow?=-2\)

\(c.\)

\(\dfrac{5}{6}\cdot\dfrac{3}{?}=\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{3}{?}=\dfrac{1}{4}:\dfrac{5}{6}=\dfrac{1}{4}\cdot\dfrac{6}{5}=\dfrac{3}{10}\)

\(\Leftrightarrow?=10\)

Mk gọi ? = x nha

a) \(\dfrac{-2}{3}.\dfrac{x}{4}=\dfrac{1}{2}\) 

           \(\dfrac{x}{4}=\dfrac{1}{2}:\dfrac{-2}{3}\) 

           \(\dfrac{x}{4}=\dfrac{-3}{4}\) 

⇒x=-3

b)\(\dfrac{x}{3}.\dfrac{5}{8}=\dfrac{-5}{12}\)  

       \(\dfrac{x}{3}=\dfrac{-5}{12}:\dfrac{5}{8}\) 

       \(\dfrac{x}{3}=\dfrac{-2}{3}\) 

⇒x=-2

c)\(\dfrac{5}{6}.\dfrac{3}{x}=\dfrac{1}{4}\) 

        \(\dfrac{3}{x}=\dfrac{1}{4}:\dfrac{5}{6}\) 

        \(\dfrac{3}{x}=\dfrac{3}{10}\) 

⇒x=10

`a)1/2 . [-3]/4 . [-5]/8 . [-8]/9=[1. (-3).(-5).(-8)]/[2.4.8.3.3]=[-5]/[2.4.3]=[-5]/24`

`b)(2/[1.3]+2/[3.5]+2/[5.7]).([10.13]/3-[2^2]/3-[5^3]/3)`

`=(1-1/3+1/3-1/5+1/5-1/7).[10.13-2^2-5^3]/3`

`=(1-1/7).[130-4-125]/3`

`=6/7 . 1/3 = 2/7`

____________________________________________________

`8/9+1/9 . 2/9+1/9 . 7/9`

`=8/9+1/9.(2/9+7/9)`

`=8/9+1/9 . 9/9`

`=8/9+1/9=9/9=1`

a) ^{_{\(\dfrac{1}{2}\cdot\dfrac{-3}{4}\cdot\dfrac{-5}{8}\cdot\dfrac{-8}{9}\)}}

^{_{\(=\dfrac{1\cdot\left(-3\right)\cdot\left(-5\right)\cdot\left(-8\right)}{2\cdot4\cdot8\cdot9}\)}}

^{_{\(=-\dfrac{5}{24}\)}}

b) ^{_{\(\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}\right)\cdot\left(\dfrac{10\cdot13}{3}-\dfrac{2^2}{3}-\dfrac{5^3}{3}\right)\)}}

^{_{\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}\right)\cdot\left(\dfrac{130}{3}-\dfrac{4}{3}-\dfrac{125}{3}\right)\)}}

_{^{\(=\left(1-\dfrac{1}{7}\right)\cdot\dfrac{1}{3}\)}}

_{^{\(=\dfrac{6}{7}\cdot\dfrac{1}{3}\)}}

_{^{\(=\dfrac{2}{7}\)}}

_{^{\(\dfrac{8}{9}+\dfrac{1}{9}\cdot\dfrac{2}{9}+\dfrac{1}{9}\cdot\dfrac{7}{9}\)}}

_{^{\(=\dfrac{8}{9}+\dfrac{2}{81}+\dfrac{7}{81}\)}}

_{^{\(=\dfrac{72}{81}+\dfrac{2}{81}+\dfrac{7}{81}\)}}

_^\(=1\)

Bài 1:

\(a,=\frac{2}{3}-\frac{16}{3}=\frac{-14}{3}\)

\(b,=\left(\frac{3}{7}+\frac{4}{7}\right)+\left(-\frac{6}{19}+\frac{-13}{19}\right)=1-1=0\)

\(c,=\frac{3}{5}.\left(\frac{8}{9}-\frac{7}{9}+\frac{26}{9}\right)=\frac{3}{5}.3=\frac{9}{5}\)

a,\(\dfrac{1}{2}\).\(\dfrac{4}{3}\)-\(\dfrac{20}{3}\).\(\dfrac{4}{5}\)=\(\dfrac{2}{3}\)-\(\dfrac{16}{3}\)=-\(\dfrac{14}{3}\)

\(\left(6-\dfrac{14}{5}\right).\dfrac{25}{8}-\dfrac{18}{5}.\dfrac{1}{4}\)

\(=\left(\dfrac{30}{5}-\dfrac{14}{5}\right).\dfrac{25}{8}-\dfrac{18}{5}.\dfrac{1}{4}\)

\(=\dfrac{16}{5}.\dfrac{25}{8}-\dfrac{18}{20}\)

\(=\dfrac{16.25}{5.8}-\dfrac{18}{20}\)

\(=\dfrac{2.5}{1}-\dfrac{9}{10}\)

\(=10-\dfrac{9}{10}\)

\(=\dfrac{100}{10}-\dfrac{9}{10}\)

\(=\dfrac{91}{10}=9\dfrac{1}{10}\)

\(\left(6-\dfrac{14}{5}\right)\) . \(\dfrac{25}{8}\) - \(\dfrac{18}{5}\) . \(\dfrac{1}{4}\)

=\(\left(\dfrac{30}{5}-\dfrac{14}{5}\right)\) . \(\dfrac{25}{8}\) - \(\dfrac{18}{5}\) .\(\dfrac{1}{4}\)

= \(\dfrac{16}{5}\) . \(\dfrac{25}{8}\) - \(\dfrac{9}{10}\)

= \(\dfrac{16.25}{5.8}\) - \(\dfrac{9}{10}\)

= \(\dfrac{2.5}{1}\) - \(\dfrac{9}{10}\)

= 10 -\(\dfrac{9}{10}\)

= \(\dfrac{100}{10}\) - \(\dfrac{9}{10}\)

= \(\dfrac{91}{10}\) = 9,1

\(=\dfrac{7}{8}\left(\dfrac{2}{9}-\dfrac{1}{18}+\dfrac{1}{36}-\dfrac{5}{12}\right)=\dfrac{7}{8}\cdot\dfrac{8-2+1-15}{36}\)

\(=\dfrac{7}{8}\cdot\dfrac{-8}{36}=\dfrac{-7}{36}\)