Hệ khá là dễ
\(\left\{{}\begin{matrix}\left(x+1\right)^2+y=xy+4\left(1\right)\\4x^2-24x+35=5\left(\sqrt{3y-11}+\sqrt{y}\right)\left(2\right)\end{matrix}\right.\)ĐKXĐ:\(y\ge\frac{11}{3}\)
\(\left(1\right)\Leftrightarrow x^2+2x+1+y-xy-4=0\)\(\Leftrightarrow\left(x-1\right)\left(x+3\right)-y\left(x-1\right)=0\)\(\Leftrightarrow\left(x-1\right)\left(x+3-y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=x+3\end{matrix}\right.\)(đến đây giải từng TH đều thay vào pt (2) bạn nhé)
\(\left\{{}\begin{matrix}\left(x+1\right)^2+y=xy+4\left(1\right)\\4x^2-24x+35=5\left(\sqrt{3y-11}+\sqrt{y}\right)\left(2\right)\end{matrix}\right.\)
\(Đkxđ:\left\{{}\begin{matrix}y\ge\frac{11}{3}\\y\ge0\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(x+1\right)^2+y=xy+4\)
\(\Leftrightarrow\left(x+1\right)^2-4=y\left(x-1\right)\)
\(\Leftrightarrow y=\frac{\left(x+1\right)^2-4}{x-1}=\frac{x^2+2x-3}{x-1}\)
\(\Leftrightarrow y=\frac{\left(x-1\right)\left(x+3\right)}{x-1}=x+3\)
\(\Rightarrow y=x+3\Rightarrow x+3\ge\frac{11}{3}\Rightarrow y\ge\frac{2}{3}\)
Thay: \(y=x+3\) vào \(\left(2\right)\) ta được:
\(4x^2-24x+35=5\left(\sqrt{3\left(x+3\right)-11}+\sqrt{x+3}\right)\)
\(\Leftrightarrow4x^2-24x+35=5\left(\sqrt{3x-2}+\sqrt{x+3}\right)\)
\(\Leftrightarrow\left(2x-7\right)\left(2x-5\right)=\frac{5\left(3x-2-x-3\right)}{\sqrt{3x-2}-\sqrt{x+3}}\)
\(\Leftrightarrow\left(2x-7\right)\left(2x-5\right)=\frac{5\left(2x-5\right)}{\sqrt{3x-2}-\sqrt{x+3}}\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-7-\frac{5}{\sqrt{3x-2}-\sqrt{x+3}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\2x-7=\frac{5}{\sqrt{3x-2}-\sqrt{x+3}}\end{matrix}\right.\)
\((*)\) \(x=\frac{5}{2}\Rightarrow y=\frac{11}{2}\left(tmđk\right)\)
\((*)\) \(2x-7=\frac{5}{\sqrt{3x-2}-\sqrt{x+3}}\)
Vì: \(x\ge\frac{2}{3}\Rightarrow2x-7-\frac{5}{\sqrt{3x-2}-\sqrt{x+3}}< 0\)
\(\Rightarrow Vô-nghiệm\)
Vậy hệ phương trình có nghiệm là: \(\left(\frac{5}{2};\frac{11}{2}\right)\)
