a/ ĐKXĐ \(x\ge4\)
\(x-1+2\sqrt{\left(x-1\right)\left(x-4\right)}+x-4=x+4\)
\(\Leftrightarrow2\sqrt{x^2-5x+4}=9-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}9-x\ge0\\4\left(x^2-5x+4\right)=\left(9-x\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\4x^2-20x+16=x^2-18x+81\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\3x^2-2x-65=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{13}{3}< 4\left(l\right)\end{matrix}\right.\)
b/ \(x^2+5x+28-5\sqrt{x^2+5x+28}-24=0\)
Đặt \(\sqrt{x^2+5x+28}=t\ge\dfrac{\sqrt{87}}{2}\) ta được:
\(t^2-5t-24=0\Rightarrow\left[{}\begin{matrix}t=8\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+5x+28}=8\Rightarrow x^2+5x-36=0\Rightarrow\left[{}\begin{matrix}x=4\\x=-9\end{matrix}\right.\)
