Question

упростите выражение

Answer 1

$\sqrt{a+1+4 \sqrt{a-3} }= \sqrt{(a+1)+ \sqrt{16(a-3)} }= \\= \sqrt{(a+1)+ \sqrt{16a-48}}=\\\\ \sqrt{ \frac{a+1+ \sqrt{(a+1)^2-(16a-48)} }{2} }+ \sqrt{ \frac{a+1- \sqrt{(a+1)^2-(16a-48)} }{2} } =\\\\= \sqrt{ \frac{a+1+ \sqrt{a^2+2a+1-16a+48} }{2} }+\sqrt{ \frac{a+1+ \sqrt{a^2+2a+1-16a+48} }{2} }=\\\\= \sqrt{ \frac{a+1+ \sqrt{a^2-14a+49} }{2} }+ \sqrt{ \frac{a+1- \sqrt{a^2-14a+49} }{2} }=\\\\= \sqrt{ \frac{a+1+ \sqrt{(a-7)^2} }{2} } +\sqrt{ \frac{a+1- \sqrt{(a-7)^2} }{2} }=$
$=\sqrt{ \frac{a+1+|a-7|}{2}} +\sqrt{ \frac{a+1-|a-7|}{2}}= \left \{ {{\sqrt{ \frac{a+1+a-7}{2}}+ \sqrt{ \frac{a+1-a+7}{2} } , a \geq 7 } \atop {\sqrt{ \frac{a+1-a+7}{2}} }+ \sqrt{ \frac{a+1+a-7}{2} } , a\ \textless \ 7} \right.\atop \right.\\\\= \left \{ {{ \sqrt{ \frac{2a-6}{2} }+2 , a \geq 7 } \atop { 2+\sqrt{ \frac{2a-6}{2} } }, a\ \textless \ 7} \right.=2+ \sqrt{a-3}$