\(\sqrt{x+1}=x-5\)
<=> \(\begin{cases}x-5\ge0\\x+1=x^2-10x+25\end{cases}\)
<=>\(\begin{cases}x\ge5\\x=\frac{11\pm\sqrt{17}}{2}\end{cases}\)
=> x=\(\frac{11+\sqrt{17}}{2}\)
vậy pt có 1 nghiệm như trên
\(\sqrt{\left(x+1\right)}=x-5\)
\(\sqrt{\left(x+1\right)^2}=\left(x-5\right)^2\)
\(\Leftrightarrow x+1=x^2-10x+25\)
\(\Leftrightarrow-x^2+11x-24=0\)
\(\Leftrightarrow-\left(x^2-11x+24\right)=0\)
\(\Leftrightarrow x^2-8x-3x+24=0\)
\(\Leftrightarrow x\left(x-8\right)-3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-8\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=8\end{cases}\left(tm\right)}\)
ĐK \(\Leftrightarrow\sqrt{x+1}\ge0\Leftrightarrow x+1\ge0\Leftrightarrow x\ge-1\)
\(\sqrt{x+1}=x-5\)
\(\Leftrightarrow\left(\sqrt{x-1}\right)^2=\left(x-5\right)^2\)
\(\Leftrightarrow x-1=x^2-10x+25\)
\(\Leftrightarrow x^2-11x+26=0\)
\(\Leftrightarrow\)\(\left[\begin{array}{nghiempt}x=\frac{11+\sqrt{17}}{2}\\x=\frac{11-\sqrt{17}}{2}\end{array}\right.\)
