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

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

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

\(\Leftrightarrow40+2xy=x\)

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

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

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

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

a) Để y nguyên thì \(6x-4⋮2x+3\)

\(\Leftrightarrow-13⋮2x+3\)

\(\Leftrightarrow2x+3\in\left\{1;-1;13;-13\right\}\)

\(\Leftrightarrow2x\in\left\{-2;-4;10;-16\right\}\)

hay \(x\in\left\{-1;-2;5;-8\right\}\)

Ta có \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\Rightarrow\left(xy+xz+yz\right)\left(x+y+z\right)=xyz\Rightarrow\left(x+y\right)\left(x+z\right)\left(y+z\right)=0\Rightarrow\left[{}\begin{matrix}x+y=0\\x+z=0\\y+z=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-y\\z=-x\\y=-z\end{matrix}\right.\)TH1: Nếu x=-y⇒x⁸-y⁸=x⁸-(-x)⁸=0 (Vì x⁸ và (-x)⁸ đều là số nguyên dương)⇒M=\(\text{​​}\dfrac{3}{4}+\left(x^8-y^8\right)\left(y^9-z^9\right)\left(z^{10}-x^{10}\right)=\dfrac{3}{4}\)

Tương tự với y=-z và z=-x

Vậy M=\(\dfrac{3}{4}\)

\(\dfrac{4}{x}-\dfrac{y}{2}=\dfrac{1}{4}\Leftrightarrow\dfrac{8-xy}{2x}=\dfrac{1}{4}\Leftrightarrow\dfrac{16-2xy}{4x}=\dfrac{x}{4x}\)

\(\Rightarrow16-2xy=x\Leftrightarrow x+2xy=16\Leftrightarrow x\left(1+2y\right)=16\)

\(\Rightarrow x;1+2y\inƯ\left(16\right)=\left\{\pm1;\pm2;\pm4;\pm8;\pm16\right\}\)

Lời giải:

$\frac{5}{x}-\frac{y}{3}=\frac{1}{6}$

$\Rightarrow \frac{15-xy}{3x}=\frac{1}{6}$

$\Rightarrow \frac{2(15-xy)}{6x}=\frac{x}{6x}$

$\Rightarrow 2(15-xy)=x$

$\Rightarrow 30=2xy+x$

$\Rightarrow 30=x(2y+1)$

$\Rightarrow x=\frac{30}{2y+1}$

Vì $x$ nguyên nên $\frac{30}{2y+1}$ nguyên

$\Rightarrow 2y+1$ là ước của $30$

Vì $2y+1$ lẻ nên $2y+1\in\left\{\pm 1; \pm 3; \pm 5; \pm 15\right\}$

$\Rightarrow y\in\left\{-1; 0; -2; 1; -3; 2; -8; 7\right\}$

Tương ứng với các giá trị $y$ trên ta có: $x\in\left\{-30; 30; -10; 10; -6; 6; -2;2\right\}$

`-1/5<=x/8<=1/4`

`=>8* -1/5<=x<=1/4*8`

`=>-8/5<=x<=2`

Mà `x in ZZ`

`=>x in {-1,0,1,2}`

Mà 

\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\) 

\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\) 

\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\) 

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