ĐK : x >= 1
<=> \(E^2=a+3+4\sqrt{a-1}-2\sqrt{\left(a+3\right)^2-16\left(a-1\right)}+a+3-4\sqrt{a-1}\)
\(=2a+6-2\sqrt{a^2+6a+9-16a+16}\)
\(=2a+6-2\sqrt{a^2-10a+25}=2a+6-2\sqrt{\left(a-5\right)^2}\)
\(=2a+6-2a+10=16\Rightarrow E=4\)
Lời giải:
\(E=\sqrt{(a-1)+4\sqrt{a-1}+4}+\sqrt{(a-1)-4\sqrt{a-1}+4}\)
\(=\sqrt{(\sqrt{a-1}+2)^2}+\sqrt{(\sqrt{a-1}-2)^2}=|\sqrt{a-1}+2|+|\sqrt{a-1}-2|\)
Nếu $a\geq 5$ thì:
\(E=\sqrt{a-1}+2+\sqrt{a-1}-2=2\sqrt{a-1}\)
Nếu $1\leq a< 5$ thì:
\(E=\sqrt{a-1}+2+(2-\sqrt{a-1})=4\)
Ta có: \(E=\sqrt{a+3+4\sqrt{a-1}}+\sqrt{a+3-4\sqrt{a-1}}\)
\(=\sqrt{a-1+2\cdot\sqrt{a-1}\cdot2+4}+\sqrt{a-1-2\cdot\sqrt{a-1}\cdot2+4}\)
\(=\left|\sqrt{a-1}+2\right|+\left|\sqrt{a-1}-2\right|\)
