Đáp án:

Giải thích các bước giải:

ĐKXĐ: $y\ne0$

Ta có:

$\begin{cases}4x^2-x+\dfrac1y=1\\y^2+y-xy^2=4\end{cases}$

$\to \begin{cases}4x^2-x+\dfrac1y=1\\1+\dfrac1y-x=\dfrac4{y^2}\end{cases}$

Đặt $\dfrac1y=t$

$\to \begin{cases}4x^2-x+t=1\\1+t-x=4t^2\end{cases}$

$\to 4x^2-x+t+t-x=4t^2$

$\to 4x^2-4t^2-2x+2t=0$

$\to 4(x-t)(x+t)-2(x-t)=0$

$\to 2(x-t)(x+t)-(x-t)=0$

$\to (x-t)(2x+2t-1)=0$

$\to x=t$ hoặc $2x+2t-1=0$

Giải: $x=t$

$\to 4x^2-x+x=1$

$\to 4x^2=1$

$\to x^2=\dfrac14$

$\to x=\dfrac{\pm1}2$

$\to \dfrac1y=t=\dfrac{\pm1}2$

$\to y=\pm2$

Giải $2x+2t-1=0$

$\to 2t=1-2x$

$\to t=\dfrac12-x$

Ta có:

$4x^2-x+t=1$

$\to 4x^2-x+\dfrac12-x=1$

$\to 4x^2-2x-\dfrac12=0$

$\to 8x^2-4x-1=0$

$\to x=\dfrac{1\pm\sqrt{3}}{4}$

$\to t\in\{\dfrac{1-\sqrt{3}}{4}, \dfrac{\sqrt{3}+1}{4}\}$

$\to \dfrac1y\in\{\dfrac{1-\sqrt{3}}{4}, \dfrac{\sqrt{3}+1}{4}\}$

$\to y\in\{\dfrac4{1-\sqrt3}, \dfrac4{1+\sqrt3}\}$