Question

Решите, пожалуйста, неравенство.

Answer 1

преобразуем неравенство

4*9/(3¹⁰/ˣ)-91/(3⁵/ˣ*4⁵/ˣ)+3*16/(4¹⁰/ˣ)≥0

умножим на 3¹⁰/ˣ, получим

36-91*(34/)⁵/ˣ+48*(3/4)¹⁰/ˣ, сделаем замену у=(3/4)⁵/ˣ>0,

36-91у+48у²≥0

36-91у+48у²=0; у=(91±√(8281-6912))/96; у=(91±37)/96; у=54/96=9/16=(3/4)²;

у=128/96=4/3=(3/4)⁻¹;

(у-(3/4)²)(у-(3/4)⁻¹)≥0

____9/16________4/3___________

+                   -                            +

а) (3/4)⁵/ˣ≤(3/4)²⇒5/х≥2; (5-2х)/х≥0 ; х=0; х=2.5;

____0_______2.5__________

-                       +                -

х∈(0;2.5]

б) (3/4)⁵/ˣ≥(3/4)⁻¹; 5/х≤-1; (5+х)/х≤0; х=0; х=-5

____-5______0_______

+               -                         +

х∈[-5;0)

Ответ х∈[-5;0)∪(0;2.5]

Answer 2

Ответ:

$\displaystyle 4\cdot 9^{1-\frac{5}{x}}-91\cdot 12^{-\frac{5}{x}}+3\cdot 4^{2-\frac{10}{x}}\geq 0$

$\star \ \ \displaystyle 4\cdot 9^{1-\frac{5}{x}}-91\cdot 12^{-\frac{5}{x}}+3\cdot 4^{2-\frac{10}{x}}=0\ \ ,\ \ \ \ ODZ:\ x\ne 0\ \ ,\\\\\\4\cdot 9\cdot 3^{-\frac{10}{x}}-91\cdot 3^{-\frac{5}{x}}\cdot 4^{-\frac{5}{x}}+3\cdot 16\cdot 4^{-\frac{10}{x}}=0\ \Big|:4^{-\frac{10}{x}}>0\\\\\\36\cdot \Big(\dfrac{3}{4}\Big)^{-\frac{10}{x}}-91\cdot \Big(\frac{3}{4}\Big)^{-\frac{5}{x}}+48=0\\\\\\t= \Big(\frac{3}{4}\Big)^{-\frac{5}{x}}>0\ \ ,\ \ \ 36t^2-91t+48=0\ \ ,\ \ \ D=1369=37^2\ \ ,$

$\displaystyle t_1=\dfrac{3}{4}\ \ ,\ \ t_2=\dfrac{16}{9}\ \ \star$

$\displaystyle \Big(t-\frac{3}{4}\Big)\Big(t-\frac{16}{9}\Big)\geq 0\ \ \ \to \ \ \ t\leq \frac{3}{4}\ \ ,\ \ t\geq \frac{16}{9}\\\\\\a)\ \ \Big(\dfrac{3}{4}\Big)^{-\frac{5}{x}}\leq \frac{3}{4}\ \ ,\ \ \ -\frac{5}{x}\geq 1\ \ ,\ \ \frac{x+5}{x}\leq 0\ \ ,\ \ x\in [-5\ ;\ 0\ )$

$\displaystyle b)\ \ \Big(\dfrac{3}{4}\Big)^{-\frac{5}{x}}\geq \frac{16}{9}\ \ ,\ \ \Big(\dfrac{3}{4}\Big)^{-\frac{5}{x}}\geq \Big(\frac{3}{4}\Big)^{-2}\ \ ,\ \ -\frac{5}{x}\leq -2\ \ ,\ \ \frac{x-2,5}{x}\leq 0\ \ ,\\\\x\in (\ 0\ ;\, 2,5\ ]\\\\\\Otvet:\ \ x\in [-5\ ;\ 0\ )\cup (\ 0\ ;2,5\ ]\ .$