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)}

ĐKXĐ: x<>0; y<>0

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

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

=>xy=-3x-12

=>xy+3x=-12

=>x(y+3)=-12

mà y+3<>-3

nên (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)}

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

\(\dfrac{x}{5}=\dfrac{y}{2}=k\)\(\Rightarrow\left\{{}\begin{matrix}x=5k\\y=2k\end{matrix}\right.\)

\(\Rightarrow xy=10k^2=1000\Rightarrow k=\pm10\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=50\\y=20\end{matrix}\right.\\\left\{{}\begin{matrix}x=-50\\y=-20\end{matrix}\right.\end{matrix}\right.\)

a: \(\dfrac{xy^2}{xy-y}=\dfrac{y\cdot xy}{y\cdot\left(x-1\right)}=\dfrac{xy}{x-1}\)

=>Hai phân thức này bằng nhau

b: \(\dfrac{xy+y}{x}=\dfrac{y\left(x+1\right)}{x}\)

\(\dfrac{xy+x}{y}=\dfrac{x\left(y+1\right)}{y}\)

Vì \(\dfrac{y\left(x+1\right)}{x}\ne\dfrac{x\left(y+1\right)}{y}\)

nên hai phân thức này không bằng nhau

c: \(\dfrac{-6}{4y}=\dfrac{-6:2}{4y:2}=\dfrac{-3}{2y}\)

\(\dfrac{3y}{-2y^2}=\dfrac{-3y}{2y^2}=\dfrac{-3y}{y\cdot2y}=\dfrac{-3}{2y}\)

Do đó: \(\dfrac{-6}{4y}=\dfrac{3y}{-2y^2}\)

=>Hai phân thức này bằng nhau

Đặt \(\dfrac{x}{5}=\dfrac{y}{2}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=5k\\y=2k\end{matrix}\right.\)\(\Rightarrow xy=10k^2\)

\(\Rightarrow k^2=1\Rightarrow k=\pm1\)

Nếu k=1 \(\Rightarrow\left\{{}\begin{matrix}x=5\\y=2\end{matrix}\right.\)

Nếu k=-1 \(\Rightarrow\left\{{}\begin{matrix}x=-5\\y=-2\end{matrix}\right.\)

\(\dfrac{xy^2}{y^2+2}=\dfrac{xy^2}{\dfrac{y^2}{2}+\dfrac{y^2}{2}+2}\le\dfrac{xy^2}{3\sqrt[3]{\dfrac{y^4}{2}}}=\dfrac{1}{3}x\sqrt[3]{2y^2}\le\dfrac{1}{9}x\left(2+y+y\right)=\dfrac{2}{9}\left(x+xy\right)\)

Tương tự: \(\dfrac{yz^2}{z^2+2}\le\dfrac{2}{9}\left(y+yz\right)\) ; \(\dfrac{zx^2}{x^2+2}\le\dfrac{2}{9}\left(z+zx\right)\)

Cộng vế:

\(P\le\dfrac{2}{9}\left(x+y+z+xy+yz+zx\right)\le\dfrac{2}{9}\left(x+y+z+\dfrac{1}{3}\left(x+y+z\right)^2\right)=4\)

Dấu "=" xảy ra khi \(x=y=z=2\)

\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)

\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\begin{array}{l}\dfrac{x}{{11}} = \dfrac{y}{{17}} = \dfrac{{x - y}}{{11 - 17}} = \dfrac{{12}}{{ - 6}} =  - 2\\ \Rightarrow x = ( - 2).11 =  - 22\\y = ( - 2).17 =  - 34\end{array}\)

Vậy \(x = -22; y = -34\).