Найти обл опр ф-ий 1) у=ln*3x+4/5-x 2) y=tg*2x-3/x+7

1y=ln[(3x+4)/(5-x)] (3x+4)/(5-x)> 0 x=-4/3   x=5           _                     +                     _   /     x∈(-4/3; 5)   2   y=tg[(2x-3)/(x+7)] -π/2< (2x-3)/(x+7)< π/2 {(2x-3)/(x+7)> -π/2   (1) {(2x-3)/(x+7)< π/2   (2) 1)(2x-3)/(x+7)+π/2> 0 (4x-6+πx+7π)/(x+7)> 0 [x(4+π)+(7π-6)]/(x+7)> 0 x=(6-7π)/(4+π)   x=-7               +                       _                                       + -7π)/(4+π)    x< -7) u x> ((6-7π)/(4+π)  ) 2)(2x-3)/(x+7)-π/2< 0 (4x-6-πx-7π)/(x+7)< 0 [x(4-π+7π)]/(x+7)< 0 x=(6+7π)/(4-π)   x=-7       +                                 _                               + +7π)/(4-π -7< x< ((6+7π)/(4-π) \\\\\\\\\\\\\                                   //////////////////////////////////////////////////////////////////////////////// -7π)/(4+π+7π)/(4-π               \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ x∈([(6-7π/(4+π)]; [(6+7π)/(4-π])