16^x+12^x-2*9^x<0 помогите пожалуйста, желательно с объяснением.

Question 1

16^x+12^x-2*9^x<0

помогите пожалуйста, желательно с объяснением.

Question 2

$16^x+12^x-2*9^x<0,$

$(4^2)^x+(4\cdot3)^x-2*(3^2)^x<0,$

$(4^x)^2+4^x\cdot3^x-2*(3^x)^2<0,$

0," alt="(4^x)^2>0," align="absmiddle" class="latex-formula">

$\frac{(4^x)^2}{(4^x)^2}+\frac{4^x\cdot3^x}{(4^x)^2}-\frac{2*(3^x)^2}{(4^x)^2}<0,$

$1+\frac{3^x}{4^x}-2(\frac{3^x}{4^x})^2<0,$

$1+(\frac{3}{4})^x-2((\frac{3}{4})^x)^2<0,$

$(\frac{3}{4})^x=a,$

0, \\ 2a^2-a-1=0," alt="1+a-2a^2<0, \\ 2a^2-a-1>0, \\ 2a^2-a-1=0," align="absmiddle" class="latex-formula">

$D=9, a_1=-\frac{1}{2}, a_2=1,$

1," alt=" [a<-\frac{1}{2}, a>1," align="absmiddle" class="latex-formula">

$[(\frac{3}{4})^x<-\frac{1}{2},$ нет решений

1," alt=" [(\frac{3}{4})^x>1," align="absmiddle" class="latex-formula">

(\frac{3}{4})^0," alt=" (\frac{3}{4})^x>(\frac{3}{4})^0," align="absmiddle" class="latex-formula">

$0<\frac{3}{4}<1,$

$x<0$

Question 3

(4^x)^2+(4^x)*(3^x)-2(3^x)^2<0

((4/3)^x)^2+(4/3)^x-2<0

(4/3)^x=v v>0

v^2+v-2<0

(v+2)(v-1)<0

v∈]-2;1[

]0;1[

0<(4/3)^x<1

x<0

arsenlevadniy_zn · Answer 1 · 2018-02-06T18:48:36+0000