Question

Решить показательные неравенства (фото)

Answer 1

решение в приложении

Answer 2

$\frac{11*3^{x-1}-31}{4*3^{2x}-11*3^{x-1}-5}-5\geq 0;\\\frac{11*3^{x-1}-31-5(4*3^{2x}-11*3^{x-1}-5)}{4*3^{2x}-11*3^{x-1}-5}\geq 0;\\\frac{11*3^{x-1}-31-180*3^{2x-2}+55*3^{x-1}+25}{36*3^{2x-2}-11*3^{x-1}-5}\geq 0;\\t=3^{x-1}; t>0;\\\frac{-180t^2+66t-6}{36t^2-11t-5}\geq 0;$

-180t^2+66t-6=0;

-30t^2+11t-1=0;

D=121-4*30=1;

t=(-11+1)/-60=1/6;

t=(-11-1)/-60=1/5;

___________

36*t^2-11t-5=0;

D=121+4*5*36=841=29²;

t=(11+29)/36=10/9;

t=(11-29)/36=-4/9;∅, т.к. t>0;

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___1/6___1/5___10/9___

t ∈ (0;1/6] ∪ [1/5;10/9);

$t=3^{x-1}=1/6;\ x-1=log_3\frac{1}{6};\ x-1=log_3\frac{1}{2}+log_3\frac{1}{3};\\x=log_3\frac{1}{2}-1+1;\ x=log_30,5;$

$t=3^{x-1}=1/5;\ x-1=log_3\frac{1}{5};\ x=log_3\frac{1}{5}+log_33;\\x=log_3\frac{3}{5}; x=log_30,6;$

$t=3^{x-1}=10/9;\ x-1=log_3\frac{10}{9};\ x=log_3\frac{10}{9}+log_33;\\x=log_3\frac{10}{3};$

x ∈ (-∞;log₃0,5] ∪ [log₃0,6;log₃10/3);

____________

7)$\frac{8*8^x-40}{2*8^{2x}-32}-1\leq 0;\ \frac{8*8^x-40-2*8^{2x}+32}{2*8^{2x}-32}\leq 0;\\\frac{8*8^x-2*8^{2x}-8}{2*8^{2x}-32}\leq 0;\ t=8^x;\ t>0;\\\frac{8t-2t^2-8}{2t^2-32}\leq 0;$

-2t²+8t-8=0;

D=64-4*2*8=0;

t=-8/-4=2;

_________

2t²-32=0;

t²=16;

t=-4;∅, т.к. t>0;

t=4;

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___2___4___

t ∈ (0;2] ∪ (4;∞);

$t=8^x=2^{3x}=2^1;\ 3x=1; x=\frac{1}{3};\\t=8^x=2^{3x}=4=2^2;\ 3x=2; x=\frac{2}{3};$

x ∈ (-∞;1/3] ∪ (2/3;∞);

8)$5^{3x}-5^{2x}+\frac{4*5^{2x}-20}{5^x-5}-4\leq 0;\\\frac{(5^x-5)(5^{3x}-5^{2x}-4)+4*5^{2x}-20}{5^x-5}\leq 0;\ t=5^x;\ t>0;\\\frac{(t-5)(t^3-t^2-4)+4t^2-20}{t-5}\leq 0;\\\frac{t^4-t^3-4t-5t^3+5t^2+20+4t^2-20}{t-5}\leq 0;\\\frac{t^4-6t^3+9t^2-4t}{t-5}\leq 0;$

t-5=0; t₁=5;

t⁴-6t³+9t²-4t=0;

t₂=0;

t³-6t²+9t-4=0;

(t³-4t²)-(2t²-8t)+(t-4)=0;

t²(t-4)-2t(t-4)+(t-4)=0;

(t-4)(t²-2t+1)=0;

t₃=4;

t²-2t+1=0;

(t-1)²=0;

t₄=1;

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__0___1___4___5___

t ∈ [1] ∪ [4;5);

$t=5^x=1=5^0;\ x=0;\\t=5^x=4;\ x=log_54;\\t=5^x=5;\ x=1;$

x ∈ [0] ∪ [log₅4;1);