a, \(\dfrac{x}{2}=-\dfrac{5}{y}\Rightarrow xy=-10\Rightarrow x;y\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

c, \(\dfrac{3}{x-1}=y+1\Rightarrow\left(y+1\right)\left(x-1\right)=3\Rightarrow x-1;y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

b: =>xy=12

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)

Áp dụng bất đẳng thức Cauchy cho hai số không âm ta có

\(x^2+\dfrac{1}{x^2}\ge2\sqrt{x^2.\dfrac{1}{x^2}}=2\)

\(y^2+\dfrac{1}{y^2}\ge2\sqrt{x^2.\dfrac{1}{x^2}}=2\)

=> \(x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge4\)

Dấu"=" xảy ra \(\Leftrightarrow x^2=\dfrac{1}{x^2};y^2=\dfrac{1}{y^2}\)

\(\Leftrightarrow x^4=1;y^4=1\Leftrightarrow x=\pm1;y=\pm1\)

Thảo ơi== Sao tao không vào hộp tin nhắn của mày với tao được==??

ĐK: x,y khác 0

Áp dụng BĐT Cô-si ta có:

\(x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}\\ \ge2\sqrt{x^2.\dfrac{1}{x^2}}+2\sqrt{y^2.\dfrac{1}{y^2}}\\ =2+2=4\)

Dấu bằng xảy ra khi và chỉ khi: \(x=y=\pm1\)

Ta có:

\(x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=4\\ \Leftrightarrow x^2-2+\dfrac{1}{x^2}+y^2-2+\dfrac{1}{y^2}=0\\ \Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)

Do \(\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\) và \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{x}\right)^2\ge0\\\left(y-\dfrac{1}{y}\right)^2\ge0\end{matrix}\right.\) nên:

\(\left(x-\dfrac{1}{x}\right)^2=\left(y-\dfrac{1}{y}\right)^2=0\)

Do đó: \(x=y=\pm1\)

d:

ĐKXĐ: y<>0; x<>0; y<>2

 \(\dfrac{4}{x}+\dfrac{2}{y}=1\)

=>\(\dfrac{4y}{xy}+\dfrac{2x}{xy}=1\)

=>2x+4y=xy

=>x(2-y)=-4y

=>x(y-2)=4y

=>\(x=\dfrac{4y}{y-2}\)

mà x,y nguyên

nên \(4y⋮y-2\)

\(\Leftrightarrow4y-8+8⋮y-2\)

=>\(y-2\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

=>\(y\in\left\{3;1;4;6;-2;10;-6\right\}\)

=>\(x\in\left\{12;-4;8;6;2;5;3\right\}\)

e: 

ĐKXĐ: x<>0; y<>0; y<>3

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{3}\)

=>\(\dfrac{x+y}{xy}=\dfrac{1}{3}\)

=>3x+3y=xy

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

=>\(x=\dfrac{3y}{y-3}\)

mà x,y nguyên

nên \(3y⋮y-3\)

=>\(3y-9+9⋮y-3\)

=>\(y-3\in\left\{1;-1;3;-3;9;-9\right\}\)

=>\(y\in\left\{4;2;6;12;-6\right\}\)

=>\(x\in\left\{12;-6;6;4;2\right\}\)

#%&$@«!?  ☺

a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)

Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)

b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)

Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)

\(\dfrac{2}{5}\) x y : \(\dfrac{7}{4}\) = \(\dfrac{7}{8}\)

\(\dfrac{2}{5}\) x y = \(\dfrac{7}{8}\) x \(\dfrac{7}{4}\)

 \(\dfrac{2}{5}\) x y = \(\dfrac{49}{32}\)

         y = \(\dfrac{49}{32}\) : \(\dfrac{2}{5}\)

         y = \(\dfrac{245}{64}\)

2\(\dfrac{2}{5}\): y x 1\(\dfrac{1}{4}\) = 2\(\dfrac{3}{5}\)

\(\dfrac{12}{5}\): y x \(\dfrac{5}{4}\) = \(\dfrac{13}{5}\)

\(\dfrac{12}{5}\): y        = \(\dfrac{13}{5}\): \(\dfrac{5}{4}\)

 \(\dfrac{12}{5}\): y = \(\dfrac{52}{25}\)

        y = \(\dfrac{12}{5}\): \(\dfrac{52}{25}\)

        y = \(\dfrac{15}{13}\)