Question

1+(1/2)log(3^1/2,(x+5)/(x+3))>=log(9,(x+1)^2)

Answer 1

$1 + frac{1}{2} log_{ sqrt{3} } { ( frac{x+5}{x+3} ) } geq log_9 {(x+1)^2}$ ;


ОДЗ:

$left{begin{array}{l} frac{x+5}{x+3} > 0 , \\ (x+1)^2 > 0 ; end{array}right$

$left{begin{array}{l} x neq -3 , \\ frac{x+5}{x+3} (x+3)^2 > 0 , \\ x neq -1 ; end{array}right$

$left{begin{array}{l} x notin { -3 , -1 } , \\ ( x + 3 ) ( x + 5 ) > 0 ; end{array}right$

$left{begin{array}{l} x notin { -3 , -1 } , \ x notin [ -5 ; -3 ] ; end{array}right$

$x notin { [ -5 ; -3 ] cup { -1 } } ;$


Решение:

$1 + log_{ sqrt{3} } { sqrt{ frac{x+5}{x+3} } } geq log_{ sqrt{9} } { sqrt{ ( x + 1 )^2 } }$ ;

$log_3 {3} + log_3 { frac{x+5}{x+3} } geq log_3 { |x+1| }$ ;

$log_3 { ( 3 cdot frac{x+5}{x+3} ) } geq log_3 { |x+1| }$ ;

$3 cdot frac{x+5}{x+3} geq |x+1|$ ;

$left[begin{array}{l} left{begin{array}{l} x < -5 , \ 3 ( x + 5 ) leq -(x+1)(x+3) ; end{array}right \\ left{begin{array}{l} -3 < x < -1 , \ 3 ( x + 5 ) geq -(x+1)(x+3) ; end{array}right \\ left{begin{array}{l} x > -1 , \ 3 (x+5) geq (x+1)(x+3) ; end{array}right end{array}right$

$left[begin{array}{l} left{begin{array}{l} x < -5 , \ 3x + 15 + x^2 + 4x + 3 leq 0 ; end{array}right \\ left{begin{array}{l} -3 < x < -1 , \ 3x + 15 + x^2 + 4x + 3 geq 0 ; end{array}right \\ left{begin{array}{l} x > -1 , \ 3x + 15 geq x^2 + 4x + 3 ; end{array}right end{array}right$

$left[begin{array}{l} left{begin{array}{l} x < -5 , \ x^2 + 7x + 18 leq 0 ; end{array}right \\ left{begin{array}{l} -3 < x < -1 , \ x^2 + 7x + 18 geq 0 ; end{array}right \\ left{begin{array}{l} x > -1 , \ x^2 + x - 12 leq 0 ; end{array}right end{array}right$

$left[begin{array}{l} left{begin{array}{l} x < -5 , \ x^2 + 2 cdot x cdot frac{7}{2} + ( frac{7}{2} )^2 + 18 - frac{49}{4} leq 0 ; end{array}right \\ left{begin{array}{l} -3 < x < -1 , \ x^2 + 2 cdot x cdot frac{7}{2} + ( frac{7}{2} )^2 + 18 - frac{49}{4} geq 0 ; end{array}right \\ left{begin{array}{l} x > -1 , \ x^2 + 2 cdot x cdot frac{1}{2} + ( frac{1}{2} )^2 - 12 - frac{1}{4} leq 0 ; end{array}right end{array}right$

$left[begin{array}{l} left{begin{array}{l} x < -5 , \ ( x + frac{7}{2} )^2 + 18 - 12 frac{1}{4} leq 0 ; end{array}right \\ left{begin{array}{l} -3 < x < -1 , \ ( x + frac{7}{2} )^2 + 18 - 12 frac{1}{4} geq 0 ; end{array}right \\ left{begin{array}{l} x > -1 , \ ( x + frac{1}{2} )^2 leq frac{49}{4} ; end{array}right end{array}right$

$left[begin{array}{l} left{begin{array}{l} x < -5 , \ ( x + frac{7}{2} )^2 leq - 5 frac{3}{4} ; end{array}right \\ left{begin{array}{l} -3 < x < -1 , \ ( x + frac{7}{2} )^2 geq - 5 frac{3}{4} ; end{array}right \\ left{begin{array}{l} x > -1 , \ | x + frac{1}{2} | leq frac{7}{2} ; end{array}right end{array}right$

$left[begin{array}{l} left{begin{array}{l} x < -5 , \ x in emptyset ; end{array}right \\ left{begin{array}{l} -3 < x < -1 , \ x in R ; end{array}right \\ left{begin{array}{l} x > -1 , \ ( x + 0.5 ) in [ -3.5 ; 3.5 ] ; end{array}right end{array}right$

$left[begin{array}{l} -3 < x < -1 , \\ left{begin{array}{l} x > -1 , \ x in [ -4 ; 3 ] ; end{array}right end{array}right$

$left[begin{array}{l} x in ( -3 ; -1 ) , \ x in ( -1 ; 3 ] ; end{array}right$

$x in ( -3 ; -1 ) cup ( -1 ; 3 ] equiv ( -3 ; 3 ]$ ${ -1 }$ ;




О т в е т : $x in ( -3 ; -1 ) cup ( -1 ; 3 ] .$

Answer 2

Не за что! :–)