log_{1/2} (x^{2}+7x+10 )>-2

log_{frac{1}{2}}(x^2+7x+10)>-2,quad ODZ:; x^2+7x+10>0\\Po; teor.; Vieta:; x_1=-2,x_2=-5\\+ + + +(-5)- - - -(-2)+ + + + +\\xin (-infty,-5)U(-2,+infty); -; ODZ\\x^2+7x+10<(frac{1}{2})^{-2}\\(frac{1}{2})^{-2}=4\\x^2+7x+6<0\\x_1=-6,x_2=-1 xin (-6,-1)\\------(-6)////(-5)--------(-2)/////(-1)----\\Otvet:; xin (-6,-5)U(-2,-1)

Предмет: Математика, автор: lordmazriel

$log_{1/2} (x^{2}+7x+10 )>-2$

Ответы

Автор ответа: NNNLLL54

$log_{frac{1}{2}}(x^2+7x+10)>-2,quad ODZ:; x^2+7x+10>0\\Po; teor.; Vieta:; x_1=-2,x_2=-5\\+ + + +(-5)- - - -(-2)+ + + + +\\xin (-infty,-5)U(-2,+infty); -; ODZ\\x^2+7x+10<(frac{1}{2})^{-2}\\(frac{1}{2})^{-2}=4\\x^2+7x+6<0\\x_1=-6,x_2=-1$

$xin (-6,-1)\\------(-6)////(-5)--------(-2)/////(-1)----\\Otvet:; xin (-6,-5)U(-2,-1)$