A>=1/(1+xy)=1/2

Dấu = xảy ra khi x=y=1

\(y\ge xy+1\ge2\sqrt{xy}\Rightarrow\sqrt{\dfrac{y}{x}}\ge2\Rightarrow\dfrac{y}{x}\ge4\)

\(Q=\dfrac{1-\dfrac{2y}{x}+2\left(\dfrac{y}{x}\right)^2}{\dfrac{y}{x}+\left(\dfrac{y}{x}\right)^2}\)

Đặt \(\dfrac{y}{x}=a\ge4\)

\(Q=\dfrac{2a^2-2a+1}{a^2+a}=\dfrac{2a^2-2a+1}{a^2+a}-\dfrac{5}{4}+\dfrac{5}{4}=\dfrac{\left(a-4\right)\left(3a-1\right)}{4\left(a^2+1\right)}+\dfrac{5}{4}\ge\dfrac{5}{4}\)

\(Q_{min}=\dfrac{5}{4}\) khi \(a=4\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

\(A=\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\left(\dfrac{1}{2xy}+8xy\right)+\dfrac{3}{xy}\)

\(A\ge\dfrac{4}{x^2+y^2+2xy}+2\sqrt{\dfrac{8xy}{2xy}}+\dfrac{3}{\dfrac{1}{4}\left(x+y\right)^2}\ge20\)

\(A_{min}=20\) khi \(x=y=\dfrac{1}{2}\)

a:

ĐKXĐ: x<>0; y<>0

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

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

=>3xy=x+2y

=>3xy-x-2y=0

=>x(3y-1)-\(2y+\frac23=\frac23\)

=>\(3x\left(y-\frac13\right)-2\left(y-\frac13\right)=\frac23\)

=>\(\left(3x-2\right)\left(y-\frac13\right)=\frac23\)

=>(3x-2)(3y-1)=2

=>(3x-2;3y-1)∈{(1;2);(2;1);(-1;-2);(-2;-1)}

=>(3x;3y)∈{(3;3);(4;2);(1;-1);(0;0)}

=>(x;y)∈{(1;1);(4/3;2/3);(1/3;-1/3);(0;0)}

mà x,y nguyên

nên x=1; y=1

b: ĐKXĐ: x<>0; y<>0

\(\frac{2}{y}-\frac{1}{x}=\frac{8}{xy}+1\)

=>\(\frac{2x-y}{xy}=\frac{8+xy}{xy}\)

=>xy+8=2x-y

=>xy-2x+y+8=0

=>x(y-2)+y-2+10=0

=>(x+1)(y-2)=-10

=>(x+1;y-2)∈{(1;-10);(-10;1);(-1;10);(10;-1);(2;-5);(-5;2);(-2;5);(5;-2)}

=>(x;y)∈{(0;-8);(-11;3);(-2;12);(9;1);(1;-3);(-6;4);(-3;7);(4;0)}

mà x<>0; y<>0

nên (x;y)∈{(-11;3);(-2;12);(9;1);(1;-3);(-6;4);(-3;7)}

d: ĐKXĐ: x<>0; y<>0

\(-\frac{3}{y}-\frac{12}{xy}=1\)

=>\(\frac{-3x-12}{xy}=1\)

=>xy=-3x-12

=>xy+3x=-12

=>x(y+3)=-12

=>(x;y+3)∈{(1;-12);(-12;1);(-1;12);(12;-1);(2;-6);(-6;2);(-2;6);(6;-2);(3;-4);(-4;3);(-3;4);(4;-3)}

=>(x;y)∈{(1;-15);(-12;-2);(-1;9);(12;-4);(2;-9);(-6;-1);(-2;3);(6;-5);(3;-7);(-4;0);(-3;1);(4;-6)}

mà y<>0

nên (x;y)∈{(1;-15);(-12;-2);(-1;9);(12;-4);(2;-9);(-6;-1);(-2;3);(6;-5);(3;-7);(-3;1);(4;-6)}

e: ĐKXĐ: y<>0

\(\frac{x}{8}-\frac{1}{y}=\frac14\)

=>\(\frac{x}{8}-\frac14=\frac{1}{y}\)

=>\(\frac{x-2}{8}=\frac{1}{y}\)

=>(x-2)y=8

=>(x-2;y)∈{(1;8);(8;1);(-1;-8);(-8;-1);(2;4);(4;2);(-2;-4);(-4;-2)}

=>(x;y)∈{(3;8);(10;1);(1;-8);(-6;-1);(4;4);(6;2);(0;-4);(-2;-2)}

mà y<>0

nên (x;y)∈{(3;8);(10;1);(1;-8);(-6;-1);(4;4);(6;2);(0;-4);(-2;-2)}

\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le\dfrac{1}{4}\)

Đặt \(\dfrac{x}{y}=a\Rightarrow0< a\le\dfrac{1}{4}\)

\(P=\dfrac{\left(\dfrac{x}{y}\right)^2-\dfrac{2x}{y}+2}{\dfrac{x}{y}+1}=\dfrac{a^2-2a+2}{a+1}=\dfrac{4a^2-8a+8}{4\left(a+1\right)}=\dfrac{4a^2-13a+3+5\left(a+1\right)}{4\left(a+1\right)}\)

\(P=\dfrac{5}{4}+\dfrac{\left(1-4a\right)\left(3-a\right)}{4\left(a+1\right)}\ge\dfrac{5}{4}\)

Dấu "=" xảy ra khi \(a=\dfrac{1}{4}\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

\(K=\left(4xy+\dfrac{1}{4xy}\right)+\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\dfrac{5}{4xy}\)

\(K\ge2\sqrt{\dfrac{4xy}{4xy}}+\dfrac{4}{x^2+y^2+2xy}+\dfrac{5}{\left(x+y\right)^2}\ge2+4+5=11\)

\(K_{min}=11\) khi \(x=y=\dfrac{1}{2}\)