ОДЗ{x+5≥0⇒x≥-5{2x-3≥0⇒x≥1,5{x-3≥0⇒x≥3x∈[3;∞)√(х+5)-√(х-3)=√(2х-3)возведем в квадратх+5+х-3-2√(х²+2х-15)>2х-32√(x²+2x-15)<54(x²+2x-15)<254x²+8x-60-25<04x²+8x-85<0D=64+1360=1424x1=(-8-4√89)/8=-1-0,5√89 U x2=-1+0,5√89 /////////////////////////////////////////////////////// + _ +-------------(-1-0,5√89)------------[3]--------(-1+0,5√89)--------------- \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\x∈[3;-1+0,5√89