ĐKXĐ: ....
Biến đổi pt dưới:
\(\Leftrightarrow x-y+2\sqrt{x-y}+1=8y-4\)
\(\Leftrightarrow\left(\sqrt{x-y}+1\right)^2=4\left(2y-1\right)\)
\(\Leftrightarrow\sqrt{x-y}+1=2\sqrt{2y-1}\) (1)
Đặt \(\left\{{}\begin{matrix}\sqrt{2y-1}=a\\\sqrt{x-y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{a^2+2b^2+1}{2}\\y=\frac{a^2+1}{2}\end{matrix}\right.\) thế vào pt trên:
\(\left(\frac{a^2+2b^2+1}{2}+\frac{3a^2+3}{2}-1\right)b+\left(a^2+2b^2+1\right)a=8\)
\(\Leftrightarrow a^3+b^3+2ab\left(a+b\right)+a+b=8\)
\(\Leftrightarrow\left(a+b\right)^3-ab\left(a+b\right)+a+b=8\) (2)
Từ (1) ta có \(b+1=2a\Rightarrow\left\{{}\begin{matrix}a+b=3a-1\\b=2a-1\end{matrix}\right.\)
Thế vào (2):
\(\left(3a-1\right)^3-\left(2a^2-a\right)\left(3a-1\right)+3a-1=8\)
\(\Leftrightarrow21a^3-22a^2+11a-10=0\)
\(\Leftrightarrow\left(a-1\right)\left(21a^2-a+10\right)=0\)
\(\Leftrightarrow a=1\Rightarrow\sqrt{2y-1}=1\Rightarrow y=1\Rightarrow x=1\)
