\(\left(x+2\right)^2-x\left(y+1\right)+y=8\)
\(\Leftrightarrow x^2+4x+4-x\left(y+1\right)+y-8=0\)
\(\Leftrightarrow x^2+x\left(3-y\right)+y-4=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-y+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=y-4\end{matrix}\right.\)
\(với:x=1\) \(thay\) \(vào\) \(pt\) \(dưới\Rightarrow15=5\sqrt{3y-14}+\sqrt{y-1}\left(đk:y\ge\dfrac{14}{3}\right)\)
\(\Leftrightarrow\)\(25\left(3y-14\right)+y-1+10\sqrt{\left(3y-14\right)\left(y-1\right)}=225\)
\(\Leftrightarrow76y-351+10\sqrt{\left(3y-14\right)\left(y-1\right)}=225\)
\(\Leftrightarrow10\sqrt{\left(3y-14\right)\left(y-1\right)}=576-76y\Leftrightarrow\left\{{}\begin{matrix}576-76y\ge0\Leftrightarrow\dfrac{14}{3}\le y\le\dfrac{144}{19}\\100\left(3y-14\right)\left(y-1\right)=\left(576-76y\right)^2\left(2\right)\end{matrix}\right.\)
giải(2) ra nghiệm y rất xấu \(y=\dfrac{21463-75\sqrt{1489}}{2783}\left(thỏa\right)\)
\(với:\)\(x=y-4\Leftrightarrow y=x+4\Rightarrow4x^2-24x+35=5\sqrt{3x-2}+\sqrt{x+3}\)
ở đây nghiệm x quá xấu lấy sấp xỉ bằng cách bình phương 2 vế nên giải bình thường
