Question

Answer 1

d: ĐKXĐ: x>=4 hoặc x<=-4

\(\sqrt{x+4}+\sqrt{x^2-16}=0\)

=>\(\sqrt{x+4}\left(\sqrt{x-4}+1\right)=0\)

=>\(\sqrt{x+4}=0\)

=>x+4=0

=>x=-4

e: ĐKXĐ:x>=0; x<>1

\(\frac{x+2\sqrt{x}}{\sqrt{x}-1}=8\)

=>\(x+2\sqrt{x}=8\left(\sqrt{x}-1\right)=8\sqrt{x}-8\)

=>\(x-6\sqrt{x}+8=0\)

=>\(\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)=0\)

=>\(\left[\begin{array}{l}\sqrt{x}-2=0\\ \sqrt{x}-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=2\\ \sqrt{x}=4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\left(nhận\right)\\ x=16\left(nhận\right)\end{array}\right.\)