\(\frac{x+1}{2}-\frac{3}{5}=\frac{1}{2y}\)

\(\Rightarrow\frac{5x+5}{10}-\frac{6}{10}=\frac{1}{2y}\)

\(\Rightarrow\frac{5x-1}{10}=\frac{1}{2y}\)

\(\Leftrightarrow\left(5x-1\right)2y=10\)

Lập bảng xong xét các trường hợp là ra

Ta có : \(\frac{x+1}{2}-\frac{3}{5}=\frac{1}{2y}\)

=> \(\frac{x+1}{2}-\frac{1}{2y}=\frac{3}{5}\)

=> \(\frac{xy+y-1}{2y}=\frac{3}{5}\)

=> 5(xy + y - 1) = 6y

=> 5xy + 5y - 5 = 6y

=> 5xy + 5y - 6y = 5

=> 5xy - y = 5

=> y(5x - 1) = 5

Vì x ; y là số nguyên

=> Ta có 5 = 1.5 = (-1).(-5)

Lập bảng xét các trường hợp

Vậy y = - 5 ; x = 0

1/

\(x^2-xy-2y^2=0\Leftrightarrow x^2+xy-2xy-2y^2=0\)

\(\Leftrightarrow x\left(x+y\right)-2y\left(x+y\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0\Rightarrow x=2y\) (do \(x+y\ne0\))

\(\Rightarrow P=\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)

2/

\(x^4-30x^2+31x-30=0\)

\(\Leftrightarrow x^4+x-30x^2+30x-30=0\)

\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)

\(\Leftrightarrow\left(x^2+x-30\right)\left(x^2-x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-30=0\\x^2-x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)\left(x+6\right)=0\\\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=5\\x=-6\end{matrix}\right.\)

\(x+y=1\Rightarrow\left\{{}\begin{matrix}y-1=-x\\x-1=-y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(y-1\right)^2=x^2\\\left(x-1\right)^2=y^2\end{matrix}\right.\)

\(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}=\frac{x}{\left(y-1\right)\left(y^2+y+1\right)}-\frac{y}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}\)

\(=\frac{-1}{y^2+y+1}+\frac{1}{x^2+x+1}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}=\frac{-1}{x^2+3y}+\frac{1}{y^2+3x}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}\)

\(=\frac{-y^2-3x+x^2+3y}{\left(xy\right)^2+3x^3+3y^3+9xy}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}=\frac{\left(x-y\right)\left(x+y\right)-3x+3y}{\left(xy\right)^2+3\left(x+y\right)\left(\left(x+y\right)^2-3xy\right)+9xy}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}\)

\(=\frac{-2\left(x-y\right)}{\left(xy\right)^2+3}+\frac{2\left(x-y\right)}{\left(xy\right)^2+3}=0\)

ta có : 1/y = x/4 - 1/2 = ( x+2)/4 <=> y = 4/(x - 2)

Để x, y nguyên nên ta có : x-2 ϵ Ư(4) = { -1 , 1 ,-2,2-4,4}

x-2=1=>x=3=>y=4

x-2=-1=>x=1=>y=-4

x-2=-2=>x=0=>y=0

x-2=2=>x=4=>y=2

x-2=-4=>x=-2=>y=-1

x-2=4=>x=6=>y=1

vay cac cap so nguyen( x,y) la :(3,4),(1,-4),(0,0),(4,2),(-2,-1),(6,1)

x4

12

1 

a) Ta có: \(\left(x-1\right)^2\ge\)0 \(\forall\)x

            \(\left|y+2\right|\ge0\)\(\forall\) y

=> \(\left(x-1\right)^2+\left|y+2\right|\ge0\)\(\forall\)x,y

=> \(\hept{\begin{cases}\left(x-1\right)^2=0\\y+2=0\end{cases}}\)

=> \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

Vậy ...

b) Ta có: \(\frac{1}{2}-\frac{y}{3}=\frac{2}{x}\)

=> \(\frac{3-2y}{6}=\frac{2}{x}\)

=> \(x\left(3-2y\right)=12\)

=> x; 3 - 2y \(\in\)Ư(12) = {1; -1; 2; -2; 3; -3; 4; -4; 6; -6; 12; -12}

Do 3 - 2y là số lẽ , mà x,y \(\in\)Z

=> 3 - 2y \(\in\) {1; -1; 3; -3} 

Lập bảng :

Vậy ...