1. đk: \(x\ge5\)
Ta có: \(PT\Leftrightarrow\sqrt{\left(x+1\right)\left(5x+9\right)}=\sqrt{\left(x+4\right)\left(x-5\right)}+5\sqrt{x+1}\)
\(\Leftrightarrow\left(x+1\right)\left(5x+9\right)=x^2+24x+5+10\sqrt{\left(x+1\right)\left(x+4\right)\left(x-5\right)}\)
\(\Leftrightarrow5x^2+14x+9-x^2-24x-5-10\sqrt{\left[\left(x+1\right)\left(x-5\right)\right]\left(x+4\right)}=0\)
\(\Leftrightarrow4x^2-10x+4-10\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}=0\)
\(\Leftrightarrow\left(2x^2-8x-10\right)+\left(3x+12\right)-5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}=0\)
\(\Leftrightarrow2\left(x^2-4x-5\right)+3\left(x+4\right)-5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}=0\)
Đặt \(\hept{\begin{cases}\sqrt{x^2-4x-5}=a\\\sqrt{x+4}=b\end{cases}}\) khi đó:
\(PT\Leftrightarrow2a^2+3b^2-5ab=0\)
\(\Leftrightarrow\left(2a^2-2ab\right)-\left(3ab-3b^2\right)=0\)
\(\Leftrightarrow2a\left(a-b\right)-3b\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a-b=0\\2a-3b=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=b\\2a=3b\end{cases}}\)
Nếu: \(a=b\Leftrightarrow\sqrt{x^2-4x-5}=\sqrt{x+4}\)
\(\Leftrightarrow x^2-4x-5=x+4\)
\(\Leftrightarrow x^2-5x-9=0\)
\(\Leftrightarrow\left(x-\frac{5+\sqrt{61}}{2}\right)\left(x-\frac{5-\sqrt{61}}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5+\sqrt{61}}{2}=0\\x-\frac{5-\sqrt{61}}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{61}}{2}\left(tm\right)\\x=\frac{5-\sqrt{61}}{2}\left(ktm\right)\end{cases}}\)
Nếu: \(2a=3b\Leftrightarrow2\sqrt{x^2-4x-5}=3\sqrt{x+4}\)
\(\Leftrightarrow4\left(x^2-4x-5\right)=9\left(x+4\right)\)
\(\Leftrightarrow4x^2-25x-56=0\)
\(\Leftrightarrow\left(x-8\right)\left(4x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=8\left(tm\right)\\x=-\frac{7}{4}\left(ktm\right)\end{cases}}\)
Vậy \(x\in\left\{\frac{5+\sqrt{61}}{2};8\right\}\)
2. đk: \(x\ge\frac{1}{2}\)
Ta có: \(x^2-2x=2\sqrt{2x-1}\)
\(\Leftrightarrow\left(x-1\right)^2-1=2\sqrt{2x-1}\)
Đặt APKHT như sau: \(a-1=\sqrt{2x-1}\)
Khi đó ta có hệ sau: \(\hept{\begin{cases}x^2-2x=2\left(y-1\right)\\y^2-2y=2\left(x-1\right)\end{cases}}\)
Trừ vế trên cho vế dưới của HPT ta được:
\(x^2-2x-y^2+2y=2\left(y-1\right)-2\left(x-1\right)\)
\(\Leftrightarrow x^2-y^2-2x+2y-2y+2x=0\)
\(\Leftrightarrow x^2-y^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)=0\)
Nếu \(x-y=0\Leftrightarrow x-1=y-1\Leftrightarrow x-1=\sqrt{2x-1}\)
\(\Leftrightarrow x^2-2x+1=2x-1\)
\(\Leftrightarrow x^2-4x+2=0\)
\(\Leftrightarrow\left(x-2-\sqrt{2}\right)\left(x-2+\sqrt{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2+\sqrt{2}\left(tm\right)\\x=2-\sqrt{2}\left(ktm\right)\end{cases}}\)
Nếu \(x+y=0\) mà \(x,y>0\) => vô lý
Vậy \(x=2+\sqrt{2}\)
