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