xy-2x+3x=5

=>xy+x=5

=>x(y+1)=5

=>(x;y+1)∈{(1;5);(5;1);(-1;-5);(-5;-1)}

=>(x;y)∈{(1;4);(5;0);(-1;-6);(-5;-2)}

\(\Leftrightarrow2x^2+x+2=y\left(2x-1\right)\)

\(\Leftrightarrow y=\dfrac{2x^2+x+2}{2x-1}=x+1+\dfrac{3}{2x-1}\)

\(y\in Z\Rightarrow\dfrac{3}{2x-1}\in Z\)

Mà x nguyên dương \(\Rightarrow2x-1>0\)

\(\Rightarrow2x-1=Ư\left(3\right)\Rightarrow x=\left\{1;2\right\}\) 

\(\Rightarrow\left(x;y\right)=\left(1;5\right);\left(2;4\right)\)

Câu a bạn giản ước đì rồi táchr a nhé

b) Ta có (x+y)²>=0

=>x²+y²+2xy>=0

=>x²+y²>= -2xy

=> x²+y²+x²+y² >=x²+y²-2xy=(x-y)²=1

=>2x²+2y²>=1

=>2x²+2y²+2>=3

=> \(\frac{2x^2+2y^2+2}{4}>=\frac{3}{4}\)

=>\(\frac{x^2+y^2+1}{2}>=\frac{3}{4}\)

Mà (x-y)²=1 => x²+y²-2xy=1

=>x²+y²-1=2xy

=.\(xy=\frac{x^2+y^2-1}{2}\) 

=> \(xy+1=\frac{x^2+y^2-1}{2}+1=\frac{x^2+y^2+1}{2}\)

=> xy+1>=3/4