⇔−840≤x40≤−540

Mà x

\(\dfrac{-1}{5}\le\dfrac{x}{40}\le\dfrac{-1}{8}\)

\(\Leftrightarrow\dfrac{-8}{40}\le\dfrac{x}{40}\le\dfrac{-5}{40}\)

\(\Leftrightarrow-8\le x\le-5\)

Mà x\(\in Z\)

\(\Rightarrow x\in\left\{-8;-7;-6;-5\right\}\)

Vậy ...

\(\dfrac{-1}{5}\le\dfrac{x}{40}< \dfrac{-1}{8}\)

<=>\(\dfrac{-8}{40}\le\dfrac{x}{40}< \dfrac{-5}{40}\)

\(\Leftrightarrow-8\le x< 5\)

Mà \(x\in Z\)

\(\Rightarrow x\in\left\{-8;-7;-6\right\}\)

Vậy

\(-\dfrac{1}{8}< \dfrac{x}{72}\le-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{-9}{72}< \dfrac{x}{72}\le-\dfrac{2}{72}\)
\(\Rightarrow x\in\left\{-8;-7;-6;-5;-4;-3;-2\right\}\)

`(-1)/8 < x/72 <= (-1)/36`

`(-1xx9)/(8xx9) < x/72 <=  (-1xx2)/(36xx2)`

`(-9)/72 < x/72 <=   (-2)/72`

`-> -9< x <=   (-2)`

`-> x=-8;-7;-6;-5;-4;-3;-2`

`@ yngoc`

a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

\(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\left(x;y\in Z\right)\)

\(MSC:8x\left(x\ne0\right)\)

\(pt\Leftrightarrow\dfrac{40+2xy}{8x}=\dfrac{x}{8x}\)

\(\Leftrightarrow40+2xy=x\)

\(\Leftrightarrow x-2xy=40\)

\(\Leftrightarrow x\left(1-2y\right)=40\)

\(\Leftrightarrow x;\left(1-2y\right)\in U\left(40\right)=\left\{-1;1;-2;2;-4;4;-5;5;-8;8;-10;10;-20;20;-40;40\right\}\)

Bạn lập bảng sẽ tìm ra các cặp \(\left(x;y\in Z\right)\) nhé!

7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\)

\(\Rightarrow\dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\)

\(\Rightarrow-2\le x\le2\)

\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)

8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)

\(\Rightarrow\dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{6}{12}-\dfrac{2}{12}\right)\)

\(\Rightarrow\dfrac{2}{3}\cdot\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}\cdot\dfrac{4}{12}\)

\(\Rightarrow\dfrac{22}{36}\le\dfrac{x}{18}\le\dfrac{28}{36}\)

\(\Rightarrow\dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\)

\(\Rightarrow x\in\left\{11;12;13;14\right\}\)

8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{3}{6}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}.\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}.\dfrac{2}{6}\\ \dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\\ \Rightarrow11\le x\le14\\ \Rightarrow x\in\left\{11;12;13;14\right\}\)

7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\\ \dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\\ \Rightarrow-2\le x\le2\\ \Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)

Đề là j zậy bn

tìm x chứ j nữa

Sao bn viết đề bài ra đi

Áp dụng BĐT Cosi:

\(\dfrac{x^2}{1+16x^4}+\dfrac{y^2}{1+16y^4}\le\dfrac{x^2}{8x^2}+\dfrac{y^2}{8y^2}=\dfrac{1}{4}\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=\pm\dfrac{1}{2}\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)

x=-1

ve trai = -1/10con ve phai=-36/35