ĐKXĐ: x<>0 và y<>0

\(\begin{cases} x^2y + 2y + x = 4xy \quad (1) \\ \dfrac{1}{x^2} + \dfrac{1}{xy} + \dfrac{x}{y} = 3 \quad (2) \end{cases}\)

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

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

=>\(\left(x+\frac{1}{y}\right)+\frac{2}{x}=4\)

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

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

Đặt \(a=x+\frac{1}{y};b=\frac{1}{x}\)

=>\(\frac{1}{y}=a-x=a-\frac{1}{b}\)

(2) sẽ trở thành: \(b^2 + \left(a - \frac{1}{b}\right)\left(b + \frac{1}{b}\right) = 3\)

=>\(b^2+a\left(b+\frac{1}{b}\right)-1-\frac{1}{b^2}=3\)

=>\(\left(b^2-\frac{1}{b^2}\right)+a\left(b+\frac{1}{b}\right)=4\)

=>\(\left(b+\frac{1}{b}\right)\left(b-\frac{1}{b}\right)+a\left(b+\frac{1}{b}\right)=4\)

=>\(\left(b+\frac{1}{b}\right)\left(b-\frac{1}{b}+a\right)=4\) (3)

\(\left(x+\frac{1}{y}\right)+\frac{2}{x}=4\)

=>a+2b=4

=>a=4-2b

=>\(b - \frac{1}{b} + (4 - 2b) = 4 - \left(b + \frac{1}{b}\right)\)

(3)=>\(\left(b + \frac{1}{b}\right) \left[4 - \left(b + \frac{1}{b}\right)\right] = 4\)

=>\(4\left(b+\frac{1}{b}\right)-\left(b+\frac{1}{b}\right)^2-4=0\)

=>\(\left(b+\frac{1}{b}\right)^2-4\cdot\left(b+\frac{1}{b}\right)+4=0\)

=>\(\left(b+\frac{1}{b}-2\right)^2=0\)

=>\(b+\frac{1}{b}-2=0\)

=>\(\frac{b^2+1-2b}{b}=0\)

=>\(\left(b-1\right)^2=0\)

=>b-1=0

=>b=1

=>\(\frac{1}{x}=1\)

=>x=1

\(a=4-2b=4-2\cdot1=4-2=2\)

\(a=x+\frac{1}{y}\)

=>\(1+\frac{1}{y}=2\)

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

=>y=1