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

khó lắm

khso vl

1) \(\hept{\begin{cases}x^2+y^2-xy=1\\x+x^2y=2y^3\end{cases}\Leftrightarrow}\hept{\begin{cases}x^2+y^2=1+xy\\x\left(1+xy\right)=2y^3\end{cases}\Rightarrow x\left(x^2+y^2\right)=2y^3}\)

\(\Leftrightarrow\left(x^3-y^3\right)+\left(xy^2-y^3\right)=0\Leftrightarrow\left(x-y\right)\left(x^2+y^2+xy\right)+y^2\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+2y^2+xy\right)=0\Leftrightarrow\orbr{\begin{cases}x=y\\x^2+2y^2+xy=0\end{cases}}\)

+) \(x=y\Rightarrow\hept{\begin{cases}y^2+y^2-y^2=1\\y+y^3=2y^3\end{cases}\Rightarrow}x=y=\pm1\)

+) \(x^2+2y^2+xy=0\)Vì y=0 không là nghiệm của hệ nên ta chia 2 vế phương trình cho y²:

\(\Rightarrow\left(\frac{x}{y}\right)^2+\frac{x}{y}+2=0\)( Vô nghiệm)

Vậy hệ có nghiệm (1;1),(-1;-1).

2/ \(\hept{\begin{cases}x+y=\sqrt{x+3y}\\x^2+y^2+xy=3\end{cases}\Rightarrow\hept{\begin{cases}x^2+y^2+2xy=x+3y\\x^2+y^2+xy=3\end{cases}}}\Rightarrow xy=x+3y-3\)

\(\Leftrightarrow\left(x-xy\right)+\left(3y-3\right)\Leftrightarrow\left(x-3\right)\left(1-y\right)=0\Leftrightarrow\orbr{\begin{cases}x=3\Rightarrow y\in\varnothing\\y=1\Rightarrow x=1\end{cases}}\)

Vậy hệ có nghiệm (1;1).

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