Question

помогите срочно пожалуйста решить систему неравенств​

Answer 1

Объяснение:

a)

$\left \{ {{5^{2x+1}>25} \atop {12^{x^2-10x}=12^{2x-20}}} \right.\ \ \ \ \left \{ {{5*5^{2x}>25\ |:5} \atop {x^2-10x=2x-20}} \right.\ \ \ \ \left \{ {{5^{2x}>5} \atop {x^2-12x+20=0}} \right. \ \ \ \ \left \{ {{2x>1\ |;2} \atop {D=64\ \ \ \ \sqrt{D}=8 }} \right.$

$\left \{ {{x>0,5} \atop {x_1=2\in\ \ x_2=10\in}} \right.$

Ответ: x₁=2    x₂=10.

б)

$\left \{ {{9^{2x-12}\geq 9^{x^2-6x}} \atop {(\frac{1}{8})^{2x+1}<(\frac{1}{64}) }} \right.\ \ \ \ \left \{ {{2x-12\geq x^2-6x} \atop {(\frac{1}{8})^{2x+1}<(\frac{1}{8})^2 }} \right.\ \ \ \ \left \{ {{x^2-8x+12\leq 0} \atop {2x+1>2}} \right. \ \ \ \ \left \{ {{x^2-6x-2x+12\leq 0} \atop {2x>1\ |:2}} \right.$

$\left \{ {{x*(x-6)-2*(x-6)\leq 0} \atop {x>0,5}} \right.\ \ \ \ \left \{ {{(x-6)*(x-2)\leq 0} \atop {x>0,5}} \right. \ \ \ \ \left \{ {{x\in[2;6]} \atop {x\in(0,5;+\infty)}} \right. \ \ \ \ \Rightarrow$

Ответ: x∈[2;6].