Bài 4:

\(a,\Rightarrow5⋮x\Rightarrow x\inƯ\left(5\right)=\left\{1;5\right\}\\ b,\Rightarrow x-2+7⋮x-2\\ \Rightarrow x-2\inƯ\left(7\right)=\left\{1;7\right\}\\ \Rightarrow x\in\left\{3;9\right\}\\ c,\Rightarrow3\left(x+1\right)+4⋮x+1\\ \Rightarrow x+1\inƯ\left(4\right)=\left\{1;2;4\right\}\\ \Rightarrow x\in\left\{0;1;3\right\}\\ d,\Rightarrow10x+6⋮2x-1\\ \Rightarrow5\left(2x-1\right)+11⋮2x-1\\ \Rightarrow2x-1\inƯ\left(11\right)=\left\{1;11\right\}\\ \Rightarrow x\in\left\{1;6\right\}\\ e,\Rightarrow x\left(x+3\right)+11⋮x+3\\ \Rightarrow x+3\inƯ\left(11\right)=\left\{1;11\right\}\\ \Rightarrow x=8\left(x\in N\right)\\ f,\Rightarrow x\left(x+3\right)+2\left(x+3\right)+5⋮x+3\\ \Rightarrow x+3\inƯ\left(5\right)=\left\{1;5\right\}\\ \Rightarrow x=2\left(x\in N\right)\)

Nhiều quá 20 câu lận

\(=\dfrac{3}{4}.\dfrac{-8}{5}+\dfrac{1}{4}=-\dfrac{6}{5}+\dfrac{1}{4}=-\dfrac{19}{20}\)

đề đâu bạn mik ko thấy

đề đâu rồi bạn ơi

\(\left(\dfrac{1}{2020}-\dfrac{2018}{2019}\right):\dfrac{2}{9}-\left(\dfrac{-2019}{2020}+\dfrac{1}{2019}\right):\dfrac{2}{3}=\)

\(\left(\dfrac{1}{2020}-\dfrac{2018}{2019}\right)\cdot\dfrac{3}{2}-\left(\dfrac{-2019}{2020}+\dfrac{1}{2019}\right)\cdot\dfrac{3}{2}=\)

\(\left[\left(\dfrac{1}{2020}-\dfrac{2018}{2019}\right)-\left(\dfrac{-2019}{2020}+\dfrac{1}{2019}\right)\right]\cdot\dfrac{3}{2}=\)

\(\left[\dfrac{1}{2020}-\dfrac{2018}{2019}-\dfrac{2019}{2020}-\dfrac{1}{2019}\right]\cdot\dfrac{3}{2}=\)

\(0\)

\(\dfrac{1}{2}-\dfrac{1}{9}:2+1=\dfrac{13}{9}\)

\(\left(\dfrac{4}{13}-\dfrac{13}{5}\right):\dfrac{4}{5}+\left(\dfrac{9}{13}+\dfrac{3}{5}\right):\dfrac{4}{5}\)

\(=\left(\dfrac{4}{13}-\dfrac{13}{5}\right)\cdot\dfrac{5}{4}+\left(\dfrac{9}{13}+\dfrac{3}{5}\right)\cdot\dfrac{5}{4}\)

\(=\left[\dfrac{4}{13}-\dfrac{13}{5}+\dfrac{9}{13}+\dfrac{3}{5}\right]\cdot\dfrac{5}{4}\)

\(=\dfrac{-5}{4}\)

Viết bài viết ngắn trang 35 là sao e =')

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