a) \(\dfrac{x}{2}=\dfrac{2}{x}\)

⇔ \(x^2=4\)

⇒ \(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

b) \(\dfrac{x}{-5}=\dfrac{-5}{x}\)

⇔ \(x^2=25\)

⇒ \(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

\(a,\Rightarrow x^2=2^2\\ \Rightarrow x=2\\ b,x^2=\left(-5\right)^2\\ \Rightarrow x=-5\)

a: =>x²=4

hay \(x\in\left\{2;-2\right\}\)

b: =>x²=25

hay \(x\in\left\{5;-5\right\}\)

\(\dfrac{x+46}{20}=x\dfrac{2}{5}\)

\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)

\(\Leftrightarrow5\cdot\left(x+46\right)=20\cdot\left(5x+2\right)\)

\(\Leftrightarrow100x+40-5x-230=0\)

\(\Leftrightarrow95x-190=0\)

\(\Leftrightarrow x=2\)

Ta có: `(x-5)/(x-3) = (x-3-2)/(x-3) = 1 - 2/(x-3)`

Để `(x-5)/(x-3)` là số nguyên thì `2/(x-3) ∈ Z`

`=> x - 3 ∈ Ư(2) = {-2;-1;1;2}`

`=> x∈ {1;2;4;5}`

Vậy `(x-5)/(x-2)` là số nguyên khi `x ∈ {-1;1;3;5}`

đề sai r

Quy đồng :

\(\dfrac{-20}{36}< \dfrac{x}{36}< \dfrac{-18}{36}\)

=> x = -19

\(-20< x< -18\Rightarrow x\in\left\{-19\right\}\)

a, \(\dfrac{x}{2}=-\dfrac{5}{y}\Rightarrow xy=-10\Rightarrow x;y\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

c, \(\dfrac{3}{x-1}=y+1\Rightarrow\left(y+1\right)\left(x-1\right)=3\Rightarrow x-1;y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

b: =>xy=12

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)

\(\dfrac{1}{2}+\dfrac{-1}{3}+\dfrac{-2}{3}\le x< \dfrac{-3}{5}+\dfrac{1}{6}+\dfrac{-2}{5}+\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{1}{2}+\left(\dfrac{-1}{3}+\dfrac{-2}{3}\right)\le x< \left(\dfrac{-3}{5}+\dfrac{-2}{5}\right)+\left(\dfrac{1}{6}+\dfrac{3}{2}\right)\)

\(\Leftrightarrow\dfrac{1}{2}+\left(-1\right)\le x< -1+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{-1}{2}\le x< \dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{-3}{6}\le x< \dfrac{4}{6}\)

\(\Leftrightarrow x\in\left\{-3;-2;-1;0;1;2;3\right\}\)

⇔x∈{−3;−2;−1;0;1;2;3}