Помогите пожалуйста найти обл опр ф-ий 1) у=ln*3x+4/5-x 2) y=tg*2x-3/x+7
Ответы на вопрос
Ответил sedinalana
0
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
{(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
+ _ +
------------------(-7)------------------((6-7π)/(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)-(6+7π)]/(x+7)<0
x=(6+7π)/(4-π) x=-7
+ _ +
--------(-7)-----------------------------((6+7π)/(4-π))--------------------
-7<x<((6+7π)/(4-π)
\\\\\\ ////////////////////////////////////////////////////////////////////////////////
---------(-7)------------------((6-7π)/(4+π))---------------------((6+7π)/(4-π))---------------------
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈([(6-7π/(4+π)];[(6+7π)/(4-π])
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
{(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
+ _ +
------------------(-7)------------------((6-7π)/(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)-(6+7π)]/(x+7)<0
x=(6+7π)/(4-π) x=-7
+ _ +
--------(-7)-----------------------------((6+7π)/(4-π))--------------------
-7<x<((6+7π)/(4-π)
\\\\\\ ////////////////////////////////////////////////////////////////////////////////
---------(-7)------------------((6-7π)/(4+π))---------------------((6+7π)/(4-π))---------------------
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈([(6-7π/(4+π)];[(6+7π)/(4-π])
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