Question

Cho x,y thỏa mãn x>1, y<0 và \(\frac{\left(x+y\right)\left(x^3-y^3\right)\sqrt{4x-2\sqrt{4x-1}}}{\left(1-\sqrt{4x-1}\right)\left(x^2y^2+xy^3+y^4\right)}=-8\). Vậy \(\frac{x}{y}=\)

Answer 1

\(A=\frac{\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\sqrt{4x-1-2\sqrt{4x-1}+1}}{-\left(\sqrt{4x-1}-1\right).y^2\left(x^2+xy+y^2\right)}=\frac{\left(x^2-y^2\right)\sqrt{\left(\sqrt{4x-1}-1\right)^2}}{-\left(\sqrt{4x-1}-1\right).y^2}\)

Do \(x>1\Rightarrow4x-1>1\Rightarrow\sqrt{4x-1}>1\Rightarrow\sqrt{4x-1}-1>0\)

\(\Rightarrow A=\frac{\left(x^2-y^2\right)\left(\sqrt{4x-1}-1\right)}{-\left(\sqrt{4x-1}-1\right).y^2}=\frac{x^2-y^2}{-y^2}=1-\left(\frac{x}{y}\right)^2\)

\(A=-8\Rightarrow1-\left(\frac{x}{y}\right)^2=-8\Rightarrow\left(\frac{x}{y}\right)^2=9\)

Do \(\left\{{}\begin{matrix}x>1\\y< 0\end{matrix}\right.\) \(\Rightarrow\frac{x}{y}< 0\Rightarrow\frac{x}{y}=-3\)