a: =-8/28-7/28=-15/28

b: \(=\dfrac{-4}{18}+\dfrac{3}{7}\cdot\dfrac{14}{15}=\dfrac{-2}{9}+\dfrac{14}{15}=\dfrac{-10+42}{45}=\dfrac{32}{45}\)

c: \(=\dfrac{-3\cdot5+7\cdot2}{20}\cdot\dfrac{-5}{1}-\dfrac{2}{9}\)

\(=\dfrac{-7}{4}-\dfrac{2}{9}=\dfrac{-63}{36}-\dfrac{8}{36}=-\dfrac{71}{36}\)

\(a,\dfrac{-2}{7}-\dfrac{1}{4}\)

\(=\dfrac{-8}{28}-\dfrac{7}{28}\)

\(=\dfrac{-15}{28}\)

\(b,\dfrac{-1}{2}.\dfrac{4}{9}+\dfrac{3}{7}\div\dfrac{15}{14}\)

a)\(\dfrac{2}{3}-\dfrac{3}{5}:\left(-1\dfrac{1}{5}\right)+\left(\dfrac{-2}{3}\right)\cdot\dfrac{3}{8}\)

\(=\dfrac{2}{3}-\dfrac{3}{5}\cdot\dfrac{-5}{6}+\left(\dfrac{-1}{4}\right)=\dfrac{5}{12}+\dfrac{1}{2}=\dfrac{11}{12}\)

b)\(17\dfrac{11}{9}-\left(6\dfrac{3}{13}+7\dfrac{11}{19}\right)+\left(10\dfrac{3}{13}-5\dfrac{1}{4}\right)=\dfrac{164}{9}-\left(\dfrac{81}{13}+\dfrac{144}{19}\right)+\left(\dfrac{133}{13}-\dfrac{21}{4}\right)=\dfrac{164}{9}-\dfrac{3411}{247}+\dfrac{259}{52}=\dfrac{6425}{684}\)

c)\(\left(\dfrac{-3}{2}\right)^2-\left[-2\dfrac{1}{3}-\left(\dfrac{3}{4}+\dfrac{1}{3}\right):2\dfrac{3}{5}\right]\cdot\left(\dfrac{-3}{4}\right)=\dfrac{9}{4}-\left[\dfrac{-7}{3}-\dfrac{13}{12}\cdot\dfrac{5}{13}\right]\cdot\left(\dfrac{-3}{4}\right)=\dfrac{9}{4}-\left(\dfrac{-11}{4}\right)\cdot\left(\dfrac{-3}{4}\right)=\dfrac{3}{16}\)

d)\(\dfrac{21}{33}:\dfrac{11}{5}-\dfrac{13}{33}:\dfrac{11}{5}+\dfrac{25}{33}:\dfrac{11}{5}+\dfrac{6}{11}=\dfrac{5}{11}\cdot\left(\dfrac{21}{33}-\dfrac{13}{33}+\dfrac{25}{33}\right)+\dfrac{6}{11}=\dfrac{5}{11}\cdot1+\dfrac{6}{11}=1\)

\(a)\dfrac{2}{3}-\dfrac{3}{5}:\left(-1\dfrac{1}{5}\right)+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)

\(=\dfrac{2}{3}-\dfrac{3}{5}:\left(\dfrac{-6}{5}\right)+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)

\(=\dfrac{2}{3}-\dfrac{-1}{2}+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)

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

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

\(=\dfrac{11}{12}\)

\(b)17\dfrac{11}{9}-\left(6\dfrac{3}{13}+7\dfrac{11}{19}\right)+\left(10\dfrac{3}{13}-5\dfrac{1}{4}\right)\)

\(=\dfrac{164}{9}-\left(\dfrac{81}{13}-\dfrac{144}{19}\right)+\left(\dfrac{133}{13}-\dfrac{21}{4}\right)\)

\(=\dfrac{164}{9}-\dfrac{-333}{247}+\dfrac{259}{52}\)

\(=19,57040036+\dfrac{259}{52}\)

\(=24,55116959\)

c)

Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)

\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)

d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

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

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

\(=3-1+\dfrac{1}{8}\)

\(=\dfrac{17}{8}\)

f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{5}{8}\)

\(=\dfrac{3}{8}+\dfrac{5}{8}\)

\(=1\)

\(\left(-\dfrac{4}{3}+\dfrac{5}{13}\right):\dfrac{2}{7}-\left(\dfrac{9}{4}+\dfrac{8}{13}\right):\dfrac{2}{7}\\ =\left(-\dfrac{4}{3}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right):\dfrac{2}{7}\\ =-\dfrac{595}{156}:\dfrac{2}{7}\\ =-\dfrac{595}{156}.\dfrac{7}{2}=-\dfrac{4165}{312}\)

\(\dfrac{-4165}{312}\)

a: Ta có: \(\left(4\sqrt{2}-\dfrac{11}{2}\sqrt{8}-\dfrac{1}{3}\sqrt{288}+\sqrt{50}\right)\cdot\left(\dfrac{1}{2}\sqrt{2}\right)\)

\(=\dfrac{1}{2}\sqrt{2}\cdot\left(4\sqrt{2}-11\sqrt{2}-4\sqrt{2}+5\sqrt{2}\right)\)

\(=\dfrac{1}{2}\sqrt{2}\cdot6\sqrt{2}=3\)

a) x: (3/4)³=(3/4)²

x = (3/4)² . (3/4)³

x = (3/4)⁵

b)(2/5)⁵ :x = (2/5)⁸

x= (2/5)⁸ : (2/5)⁵

x= (2/5)³

a, \(x:\left(\dfrac{3}{4}\right)^3=\left(\dfrac{3}{4}\right)^2\)

=> \(x=\left(\dfrac{3}{4}\right)^2.\left(\dfrac{3}{4}\right)^3\)

=> \(x=\left(\dfrac{3}{4}\right)^5\)

b, \(\left(\dfrac{2}{5}\right)^5:x=\left(\dfrac{2}{5}\right)^8\)

\(x=\left(\dfrac{2}{5}\right)^5:\left(\dfrac{2}{5}\right)^8\)

\(x=\left(\dfrac{2}{5}\right)^{-3}\)

a) \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)

⇔ \(\left(\dfrac{-8}{5}+x\right).\dfrac{13}{12}=\dfrac{13}{6}\)

⇔ \(-\dfrac{8}{5}+x=\dfrac{13}{6}:\dfrac{13}{12}\)

⇔ \(-\dfrac{8}{5}+x=2\)

⇔ \(x=2+\dfrac{8}{5}\)

⇔ \(x=\dfrac{18}{5}\)

b) \(\dfrac{-4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)

⇔ \(-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)

⇔ \(-\dfrac{4}{7}x=-\dfrac{3}{40}-\dfrac{7}{5}\)

⇔ \(-\dfrac{4}{7}x=-\dfrac{59}{40}\)

⇔ \(x=\left(-\dfrac{59}{40}\right):\left(-\dfrac{4}{7}\right)\)

⇔ \(x=\dfrac{413}{160}\)

a, \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)

=> \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=\dfrac{13}{6}\)

=> \(\left(-1\dfrac{3}{5}+x\right)=\dfrac{13}{6}.\dfrac{12}{13}\)

=> \(\left(-1\dfrac{3}{5}+x\right)=2\)

=> \(\dfrac{-8}{5}+x=2\)

=> x= \(2+\dfrac{8}{5}=\dfrac{10}{5}+\dfrac{8}{5}\)

=> x= \(\dfrac{18}{5}\)

\(a,x=\dfrac{1}{5}+\dfrac{-3}{7}\)

   \(x=\dfrac{7}{35}+\dfrac{-15}{35}\)

   \(x=-\dfrac{8}{35}\)

\(b,\dfrac{3}{5}-\dfrac{4}{7}:x=\dfrac{-9}{10}\)

           \(\dfrac{4}{7}:x=\dfrac{3}{5}-\dfrac{-9}{10}\)

           \(\dfrac{4}{7}:x=\dfrac{3}{2}\)

                 \(x=\dfrac{4}{7}:\dfrac{3}{2}\)

                 \(x=\dfrac{4}{7}\times\dfrac{2}{3}\)

                 \(x=\dfrac{8}{21}\)

\(c,x-\left(\dfrac{-3}{4}\right)=\dfrac{-2}{3}-\dfrac{1}{2}\)

   \(x+\dfrac{3}{4}=\dfrac{-4}{6}-\dfrac{3}{6}\)

   \(x+\dfrac{3}{4}=-\dfrac{7}{6}\)

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

           \(x=-\dfrac{23}{12}\)

\(d,\dfrac{-5}{9}-x=\dfrac{1}{3}+\dfrac{7}{18}\)

    \(\dfrac{-5}{9}-x=\dfrac{6}{18}+\dfrac{7}{18}\)

     \(\dfrac{-5}{9}-x=\dfrac{13}{18}\)

                \(x=\dfrac{-5}{9}-\dfrac{13}{18}\)

                \(x=\dfrac{-10}{18}-\dfrac{13}{18}\)

                \(x=-\dfrac{23}{18}\)

giúp mk vs ạ

`a)sqrt{28a^4}`

`=sqrt{7.4.a^4}`

`=2sqrt7a^2`

`b)A=((sqrt{21}-sqrt7)/(sqrt3-1)+(sqrt{10}-sqrt5)/(sqrt2-1)):1/(sqrt7-sqrt5)`

`=((sqrt7(sqrt3-1))/(sqrt3-1)+(sqrt5(sqrt2-1))/(sqrt2-1)).(sqrt7-sqrt5)`

`=(sqrt7+sqrt5)(sqrt7-sqrt5)`

`=7-5=2`

`c)` $\begin{cases}\dfrac{3}{2x}-y=6\\\dfrac{1}{x}+2y=-4\end{cases}$

`<=>` $\begin{cases}\dfrac{3}{x}-2y=12\\\dfrac{1}{x}+2y=-4\end{cases}$

`<=>` $\begin{cases}\dfrac{4}{x}=8\\2y+\dfrac{1}{x}=-4\end{cases}$

`<=>` $\begin{cases}x=\dfrac12\\2y=-4-2=-6\end{cases}$

`<=>` $\begin{cases}x=\dfrac12\\y=-3\end{cases}$

Vậy HPT có nghiệm `(x,y)=(1/2,-3)`.