Question

Помогите решить уравнения

Answer 1

$\displaystyle 2^{x^2-1}-3^{x^2}=3^{x^2-1}-2^{x^2+2}\\ 2^{x^2-1}+2^{x^2+2}=3^{x^2-1}+3^{x^2}\\ 2^{x^2-1}(1+8)=3^{x^2-1}(1+3)\\ 2^{x^2-1}\cdot 9=3^{x^2-1}\cdot 4 \;\;\;\;\;\;\;\;\;\;\;\;\;\; |:36\\ 2^{x^2-3}=3^{x^2-3} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;|: 3^{x^2-3}\\\\ (\frac{2}{3})^{x^2-3}=1\\\\ (\frac{2}{3})^{x^2-3}=(\frac{2}{3})^0\\\\ x^2-3=0\\x^2=3\\\\ \left[{ {{x=\sqrt{3}} \atop {x=-\sqrt{3}} \right.$

Ответ: $-\sqrt{3};\sqrt{3}$

$\displaystyle 3\cdot 5^{2x-1}-2\cdot 5^{x-1}=0,2\;\;\;\;\;\;\;\;\;\; |\cdot 5 \\ 3\cdot 5^{2x}-2\cdot 5^x-1=0\\ 5^x=y; \;\;\;\;y>0\\3y^2-2y-1=0 \\ D=4+12=16\\ y_{1,2}=\frac{2б4}{6}=\left |{ {{1} \atop {-\frac{1}{3}}} \right. \\\\y=1 \;\;\;\; \Leftrightarrow \;\;\;\; 5^x=1\\x=0$

Ответ: $0$

$\displaystyle 9^{x^2-1}-36\cdot 3^{x^2-3}+3=0\\3^{2(x^2-1)}-4\cdot 3^{x^2-1}+3=0\\y=3^{x^2-1}; \;\;\;\; y>0\\ y^2-4y+3=0\\ (y-3)(y-1)=0\\\left [{ {{y=3} \atop {y=1}} \right. \left [{ {{3^{x^2-1}=3} \atop {3^{x^2-1}=1} \right. \left [{ {{x^2-1=1} \atop {x^2-1=0}} \right. \left [{ {{\left { {{x=\sqrt{2}} \atop {x=-\sqrt{2}}} \right. } \atop {\left { {{x=1} \atop {x=-1}} \right. }} \right.$

Ответ: $-\sqrt{2};-1;1;\sqrt{2}$