\(\Leftrightarrow\dfrac{1}{a}=\dfrac{2b-3}{4}\Rightarrow a=\dfrac{4}{2b-3}\left(b\ne\dfrac{3}{2}\right)\) (1)

\(a\in Z\Rightarrow\left(2b-3\right)=\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow b=\left\{-\dfrac{1}{2};\dfrac{1}{2};1;2;\dfrac{5}{2};\dfrac{7}{2}\right\}\) Do \(b\in Z\Rightarrow b=\left\{1;2\right\}\)

Thay vào (1) \(\Rightarrow a=\left\{-4;4\right\}\)

Ko có gt thỏa mãn

-12

-3/5 = x/20
-12/20 = x/20
x/20 = -12/20
x/20 = -12 . 20
x/20 = -240
x      = -240 : 20
x      = -12
Vậy số nguyên x dể thỏa mãn đề bài là -12.

-12

\(PT\Leftrightarrow\dfrac{5\left(a-b\sqrt{2}\right)}{a^2-2b^2}-\dfrac{4\left(a+b\sqrt{2}\right)}{a^2-2b^2}+18\sqrt{2}-3=0\\ \Leftrightarrow\left(\dfrac{5a}{a^2-2b^2}-\dfrac{4a}{a^2-2b^2}-3\right)+\left(18\sqrt{2}-\dfrac{5b\sqrt{2}}{a^2-2b^2}-\dfrac{4b\sqrt{2}}{a^2-2b^2}\right)=0\\ \Leftrightarrow\left(\dfrac{5a}{a^2-2b^2}-\dfrac{4a}{a^2-2b^2}-3\right)+\sqrt{2}\left(18-\dfrac{5b}{a^2-2b^2}-\dfrac{4b}{a^2-2b^2}\right)=0\)

Vì a,b nguyên mà vế trái có \(\sqrt{2}\) vô tỉ nên 2 biểu thức còn lại phải bằng 0

 \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5a}{a^2-2b^2}-\dfrac{4a}{a^2-2b^2}=3\\\dfrac{5b}{a^2-2b^2}+\dfrac{4b}{a^2-2b^2}=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{a^2-2b^2}=3\\\dfrac{b}{a^2-2b^2}=2\end{matrix}\right.\left(a,b\ne0\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-2b^2=\dfrac{a}{3}\\b=2\left(a^2-2b^2\right)=2\cdot\dfrac{a}{3}=\dfrac{2}{3}a\end{matrix}\right.\)

\(\Leftrightarrow a^2-\dfrac{8}{9}a^2=\dfrac{a}{3}\Leftrightarrow\dfrac{1}{9}a^2-\dfrac{1}{3}a=0\Leftrightarrow\dfrac{1}{3}a\left(\dfrac{1}{3}a-1\right)=0\\ \Leftrightarrow a=3\left(a\ne0\right)\)

\(\Leftrightarrow b=\dfrac{2}{3}\cdot3=2\left(tm\right)\)

Vậy \(\left(a;b\right)=\left(3;2\right)\)

\(\dfrac{x}{8}-\dfrac{1}{4}=\dfrac{1}{y}\)

\(\Leftrightarrow\dfrac{x-2}{8}=\dfrac{1}{y}\)

\(\Leftrightarrow x-2=\dfrac{8}{y}\)

Do \(x-2\in Z\Rightarrow\dfrac{8}{y}\in Z\)

\(\Rightarrow y=Ư\left(8\right)\)

\(\Rightarrow y=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

\(\Rightarrow x=\left\{1;0;-2;-6;10;6;4;3\right\}\)

=>-3y+15=14-2y

=>-y=-1

=>y=1

\(\dfrac{1}{3}< \dfrac{x}{y}< \dfrac{1}{2}\Rightarrow\dfrac{4}{12}< \dfrac{x}{y}< \dfrac{6}{12}\Rightarrow\dfrac{x}{y}=\dfrac{5}{12}\Rightarrow\dfrac{x}{5}=\dfrac{y}{12}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{5}=\dfrac{y}{12}=\dfrac{2x}{10}=\dfrac{3y}{36}=\dfrac{2x+3y}{10+36}=\dfrac{19}{46}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{95}{46}\\y=\dfrac{114}{23}\end{matrix}\right.\)

Mà \(x,y\in Z\)

Vậy ko có x,y nguyên thỏa mãn đề

khó phết

a: =>-2x=90/91

hay x=-45/91

b: =>2x=-7

hay x=-7/2

c: ->-3x=-12

hay x=4