4^1/x + 6^1/x - 9^1/x = 0 Помогите решить срочно

Question 1

Question 2

(4) ^(1/x) + (6) ^(1/x) - (9)^(1/x) =0 ⇔ (2) ^(2/x) +(2) ^(1/x)*(3)^(1/x) - (3)^(2/x) =0

Question 3

однородное уравнение второго порядка

Question 4

0\; ,\; \; t^2+t-1=0\; ,\; \; (t+0,5)^2-1,25=0\; ," alt="4^{1/x}+6^{1/x}-9^{1/x}=0\; ,\; \; x\ne 0\\\\(2^{1/x})^2+2^{1/x}\cdot 3^{1/x}-(3^{1/x})^2=0\; |:(3^{1/x})^2\ne 0\\\\t=(\frac{2}{3})^{1/x}>0\; ,\; \; t^2+t-1=0\; ,\; \; (t+0,5)^2-1,25=0\; ," align="absmiddle" class="latex-formula">

0\; \; ,\; \; t_2=-0,5-\sqrt{1,25}<0\; \; ne\; podxodit" alt="(t+0,5-\sqrt{1,25})(t+0,5+\sqrt{1,25})=0\\\\t_1=\sqrt{1,25}-0,5>0\; \; ,\; \; t_2=-0,5-\sqrt{1,25}<0\; \; ne\; podxodit" align="absmiddle" class="latex-formula">

$(\frac{2}{3})^{1/x}=\sqrt{\frac{125}{100}}-\frac{1}{2}\; ,\; \; (\frac{2}{3})^{1/x}=\frac{5\sqrt5}{10}-\frac{1}{2}=\frac{\sqrt5-1}{2}$

$\frac{1}{x}=log_{2/3}\frac{\sqrt5-1}{2}\; ,\; \; x=\frac{1}{log_{2/3}\frac{\sqrt5-1}{2}}$

$x=\frac{log_2\frac{2}{3}}{log_2(\sqrt{5}-1)-log_22}=\frac{1-log_23}{log_2(\sqrt5-1)-1}$

NNNLLL54_zn · Answer 1 · 2020-05-17T23:15:56+0000

0\; ,\; \; t^2+t-1=0\; ,\; \; (t+0,5)^2-1,25=0\; ," alt="4^{1/x}+6^{1/x}-9^{1/x}=0\; ,\; \; x\ne 0\\\\(2^{1/x})^2+2^{1/x}\cdot 3^{1/x}-(3^{1/x})^2=0\; |:(3^{1/x})^2\ne 0\\\\t=(\frac{2}{3})^{1/x}>0\; ,\; \; t^2+t-1=0\; ,\; \; (t+0,5)^2-1,25=0\; ," align="absmiddle" class="latex-formula">

0\; \; ,\; \; t_2=-0,5-\sqrt{1,25}<0\; \; ne\; podxodit" alt="(t+0,5-\sqrt{1,25})(t+0,5+\sqrt{1,25})=0\\\\t_1=\sqrt{1,25}-0,5>0\; \; ,\; \; t_2=-0,5-\sqrt{1,25}<0\; \; ne\; podxodit" align="absmiddle" class="latex-formula">

$(\frac{2}{3})^{1/x}=\sqrt{\frac{125}{100}}-\frac{1}{2}\; ,\; \; (\frac{2}{3})^{1/x}=\frac{5\sqrt5}{10}-\frac{1}{2}=\frac{\sqrt5-1}{2}$

$\frac{1}{x}=log_{2/3}\frac{\sqrt5-1}{2}\; ,\; \; x=\frac{1}{log_{2/3}\frac{\sqrt5-1}{2}}$

$x=\frac{log_2\frac{2}{3}}{log_2(\sqrt{5}-1)-log_22}=\frac{1-log_23}{log_2(\sqrt5-1)-1}$