ĐK: x \(\ge\)-1 . pt \(\Leftrightarrow\) (x+1 -4\(\sqrt{x+1}\) +4) + \(^{x^2}\) -6x+9 =0 \(\Leftrightarrow\) (\(\sqrt{x+1}\) -2)^2 +(x-3 )^2 =0 Do (\(\sqrt{x+1}\)-2)^2 \(\ge\)0 ; (x-3)^2 \(\ge\)0 \(\Rightarrow\) \(\left\{{}\begin{matrix}\sqrt{x+1}=2\\x-3=0\end{matrix}\right.\)\(\Rightarrow\) x=3 (tm)
Đk:\(x\ge-1\)
\(pt\Leftrightarrow4\sqrt{x+1}-8=x^2-5x+6\)
\(\Leftrightarrow4\left(\sqrt{x+1}-2\right)=x^2-5x+6\)
\(\Leftrightarrow\dfrac{4\left(x-3\right)}{\left(\sqrt{x+1}-2\right)}=\left(x-2\right)\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(\dfrac{4}{\sqrt{x+1}-2}-x+2\right)=0\)
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\) (thỏa mãn)