Ответы на вопрос:
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])
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