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

Question

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

Comments

( * ) - знак умножения. Нельзя его ставить после названия функции.

Answer 1

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