Урок 18. Показательные уравнения, неравенства и системы
Домашнее задание из В.В. Ткачук “Математика – абитуриенту”
- $$3\cdot 4^x-7\cdot 10^x+2\cdot 25^x=0$$
- $$3^x-3^{\frac{1}{2}-x}>\sqrt{3}-1$$
- $$2^{3x-\frac{1}{2}}+\frac{\sqrt{2}}{2}>1+2^{-3x}$$
- $$32^{3(x^3-8)}=8^{19(2x-x^2)}$$
- $$\displaystyle\frac{5^x}{2^{x-1}-5^x}=8-\displaystyle\frac{2^{x+1}}{5^x}$$
- $$7^x-14\cdot 7^{-x}=3^{\log_32+3}$$
- \(\left\{\begin{array}{l l} 6^x-2\cdot 3^y=2,\\6^x\cdot 3^y=12\end{array}\right.\)
- $$(\displaystyle\frac{1}{4})^{\displaystyle\frac{4-x^2}{2}}=8^x$$
- $$(\displaystyle\frac{1}{8})^{\displaystyle\frac{2x^2}{3}}=4^{-x}\cdot 8^{-4}$$
- $$7^{x-\frac{x^2}{8}}<7^{1-x}\cdot (\sqrt[\displaystyle 8]{7})^{x^2}+6$$
- $$5^{2x-\frac{x^2}{3}}<5^{2-2x}\cdot (\sqrt[\displaystyle 3]{5})^{x^2}+24$$
- $$3^x<1+12\cdot 3^{-x}$$
- \(\left\{\begin{array}{l l} 7\cdot 2^x+6y=2,\\3\cdot 2^{x+1}-5y=93\end{array}\right.\)
- $$\displaystyle\frac{2\cdot 6^x-4^x-15}{6^x-9^x-5}=3$$
- $$\displaystyle\frac{15^x+9^x+6}{2\cdot 15^x+25^x+3}=2$$
- $$5^{x+1}=(\frac{1}{5})^{x-2}$$
- $$4\cdot 4^x<7\cdot 2^x+2$$
- [2] $$2^{x+3}-x^3\cdot 2^x\leq 16-2x^3$$
- [2] $$x^4+3^{x+4}\geq x^4\cdot 3^x+81$$
- $$9^x-6\cdot 3^x-27=0$$
- $$2\cdot 4^{x+2}+14\cdot 2^x-1=0$$
- $$4^{-\displaystyle\frac{1}{x}}+6^{-\displaystyle\frac{1}{x}}=9^{-\displaystyle\frac{1}{x}}$$
- $$4^x\leq 3\cdot 2^{\sqrt{x}+x}+4^{1+\sqrt{4}}$$
- $$\displaystyle\frac{1}{2^x-1}>\displaystyle\frac{1}{1-2^{x-1}}$$
- $$(x^2-x+1)^x<1$$
- $$(x^2+x+1)^x<1$$
- $$(\sqrt{2-\sqrt{3}})^x+(\sqrt{2+\sqrt{3}})^x=4$$
- $$2^{x+3}-3^{x^2+2x-6}=3^{x^2+2x-5}-2^x$$
- $$4^{x^2}-3\cdot 2^{x^2}+1\geq 0$$
- $$2\cdot (\frac{7^x+7^{-x}}{2})^2-7\cdot \frac{7^x+7^{-x}}{2}+3=0$$
Ответы к домашнему заданию урока 18 из В.В. Ткачук “Математика – абитуриенту”
- $$\log_{2/5}2; \log_{2/5}(1/3)$$
- $$(1/2; +\infty)$$
- $$(1/6; +\infty)$$
- 2; -5; -4/5
- $$\log_{2/5}3$$
- 1
- $$(1; \log_32)$$
- 4; -1
- 3; -2
- $$(-\infty; 4-2\sqrt{2})\cup (4+2\sqrt{2}; +\infty)$$
- $$(-\infty; 3-\sqrt{3})\cup (3+\sqrt{3}; +\infty)$$
- $$(-\infty; \log_34)$$
- (3;-9)
- $$\log_{3/2}((\sqrt{13}+1)/6)$$
- $$\log_{5/3}((\sqrt{17}-3)/4)$$
- 1/2
- $$(-\infty; 1)$$
- $$(-\infty; 1]\cup [2; +\infty)$$
- $$(-\infty; -3]\cup [0;3]$$
- 2
- -4
- $$1/\log_{3/2}((\sqrt{5}-1)/2)$$
- [0;4]
- $$(0; 2-\log_23)\cup (1; +\infty)$$
- $$(-\infty; 0)\cup (0;1)$$
- $$(-\infty; -1)$$
- 2; -2
- 2; $$\log_32-4$$
- $$(-\infty; -\sqrt{\log_2(3+\sqrt{5})-1}]\cup[\sqrt{\log_2(3+\sqrt{5})-1};+\infty)$$
- $$\log_7(3\pm\sqrt{2})$$