Question

Решите неравенства все , много ​

Answer 1

$1)\; \; 3^{x+1}+3^{x+2}+3^{x+3}4^{x}+4^{x+1}\\\\3^{x}\cdot (3+3^2+3^3)4^{x}\cdot (1+4)\\\\3^{x}\cdot 394^{x}\cdot 5\; \; \Rightarrow \; \; \frac{3^{x}}{4^{x}}\frac{5}{39}\; \; ,\; \; (\frac{3}{4})^{x}\frac{5}{39}\; \; \Rightarrow \; \; x<log_{0,75}\frac{5}{39}\\\\\\2)\; \; 9^{2x^2-5x}+3^{2x^2-5x+1}-40\\\\(3^{2x^2-5x})^2+3^{2x^2-5x}\cdot 3-40\\\\t=3^{2x^2-5x}0\; \; ,\; \; t^2+3t-40\; ,\; \; t_1=-4\; ,\; t_2=1\; \; (Viet)\\\\znaki:\; \; \; +++(-4)---(1)+++\; \; \;\qquad t1\; \; ili\; \; t<-4$

$a)\; \; 3^{2x^2-5x}1\; \; ,\; \; 3^{2x^2-5x}3^0\; \; \to \; \; 2x^2-5x0\; ,\\\\x(2x-5)0\; \; ,\; \; x_1=0\; ,\; x_2=2,5\\\\+++(0)---(2,5)+++\\\\x\in (-\infty ;0)\cup (2,5\, ;\, +\infty )\\\\b)\; \; 3^{2x^2-5x}<-4\; \; \Rightarrow \; \; x\in \varnothing\; ,\; t.k.\; \; 3^{2x^2-5x}0\; .\\\\Otvet:\; \; x\in (-\infty \, ;0)\cup (2,5\, ;+\infty )\; .$

$3)\; \; 3\cdot 16^{x}+2\cdot 81^{x}\leq 5\cdot 36^{x}\\\\3\cdot (2^{x})^4+2\cdot (3^{x})^4-5\cdot (6^{x})^2\leq 0\; |:(6^{x})^2\\\\3\cdot (\frac{2^{x}}{3^{x}})^2+2\xsot (\frac{3^{x}}{2^{x}})^2-5\leq 0\\\\t=(\frac{2^{x}}{3^{x}})^20\; \; ,\; \; \; 3t+\frac{2}{t}-5\leq 0\; ,\; \; \; \frac{3t^2-5t+2}{t}\leq 0\; ,\; \; 3t^2-5t+2\leq 0\; ,\\\\D=1\; ,\; t_1=\frac{2}{3}\; \; ,\; \; t_2=1\\\\znaki:\; \; \; +++[\, \frac{2}{3}\, ]---[\, 1\, ]+++\; \; \; \quad t\in [\, \frac{2}{3}\, ;\, 1\, ]$

$(\frac{2^{x}}{3^{x}})^{2}=(\frac{2}{3})^{2x}\; \; ,\; \; \; \; \frac{2}{3}\leq (\frac{2}{3})^{2x}\leq 1\; \; \Rightarrow \; \; 0\leq 2x\leq 1\; \; ,\; \; 0\leq x\leq 0,5\\\\Otvet:\; \; x\in [\, 0\, ;\, 0,5\, ]\; .$

$4)\; \; (x-2)^{x^2-6x+8}1\\\\(x^2-6x+8)\cdot lg(x-2)lg1\qquad [\; x^2-6x+8=0\; \to \; x_1=2\; ,\; x_2=4\, ]\\\\(x^2-6x+8)\cdot lg(x-2)0\\\\\left \{ {{x^2-6x+80} \atop {lg(x-2)0}} \right. \; \; ili\; \; \left \{ {{x^2-6x+8<0} \atop {lg(x-2)<0}} \right. \\\\\left \{ {{(x-2)(x-4)0} \atop {x-21}} \right. \; \; ili\; \; \left \{ {{(x-2)(x-4)<0} \atop {x-2<1}} \right.\\\\\left \{ {{x\in (-\infty ,2)\cup (4,+\infty )} \atop {x3}} \right.\; \; ili\; \; \left \{ {{x\in (2,4)} \atop {x<3}} \right.$

$x\in (4,+\infty )\; \; \quad ili\; \; \quad \; \; x\in (2,3)\\\\Otvet:\; \; x\in (2,3)\cup (4,+\infty )\; .$