a: Sửa đề: \(\sqrt{x+2\sqrt{x+1}}=2\)
ĐKXĐ: x>=-1
Ta có: \(\sqrt{x+2\sqrt{x+1}}=2\)
=>\(x+2\sqrt{x+1}=4\)
=>\(x+1+2\sqrt{x+1}+1=4+2=6\)
=>\(\left(\sqrt{x+1}+1\right)^2=6\)
=>\(\sqrt{x+1}+1=\sqrt6\) (Vì \(\sqrt{x+1}+1\ge1\forall x\) thỏa mãn ĐKXĐ)
=>\(\sqrt{x+1}=\sqrt6-1\)
=>\(x+1=\left(\sqrt6-1\right)^2=7-2\sqrt6\)
=>\(x=6-2\sqrt6\) (nhận)
b: Sửa đề: \(\sqrt{x+4\sqrt{x-4}}=2\)
ĐKXĐ: x>=4
Ta có: \(\sqrt{x+4\sqrt{x-4}}=2\)
=>\(\sqrt{x-4+4\sqrt{x-4}+4}=2\)
=>\(\sqrt{\left(\sqrt{x-4}-2\right)^2}=2\)
=>\(\left|\sqrt{x-4}-2\right|=2\)
=>\(\left[\begin{array}{l}\sqrt{x-4}-2=-2\\ \sqrt{x-4}-2=2\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x-4}=0\\ \sqrt{x-4}=4\end{array}\right.\Rightarrow\left[\begin{array}{l}x-4=0\\ x-4=16\end{array}\right.\)
=>\(\left[\begin{array}{l}x=4\left(nhận\right)\\ x=20\left(nhận\right)\end{array}\right.\)
