Question

determinati valorile admisibile ale variabilelor $\alpha$ si aduceti la forma ce mai simpla expresia :

$(2^{log\sqrt[4]{2} ^{a} } -3^{log27^{} {(a^{2} +1)^{3} } } } -2a):(7^{4log49^{a} } -a-1)$

va-implooooor

Answer 1

$\text{Formule:}\\\\ a^{\log_{a}x} = x\\ \log_{a^x}b = \log_{a}b^{\frac{1}{x}}\\ \log_{a}b^x = \log_{a^{\frac{1}{x}}}b\\\\\\\left(2^{\log_{\sqrt[4]{2}}a}-3^{\log_{27}(a^2+1)^3} - 2a\right):\left(7^{4\log_{49}a}-a-1\right) = \\ \\ = \left(2^{\log_{2}a^4}-3^{\log_{3^3}(a^2+1)^3}-2a\right):\left(7^{\log_{7^2}a^{4}}-a-1\right) = \\ \\ =\left(a^4-3^{\log_{3}(a^2+1)}-2a\right):\left(7^{\log_{7}a^2}-a-1\right) = \\ \\ = \left(a^4-(a^2+1)-2a\right):(a^2-a-1)=$

$= \left(a^4-(a^2+2a+1)\right):\left(a^2-a-1\right) = \\ \\ = \left((a^2)^2-(a+1)^2\right):\left(a^2-a-1\right) =\\ \\ =\left(a^2-(a+1)\right)\left(a^2+a+1\right):\left(a^2-a-1\right) = \\ \\ = \left(a^2-a-1\right)\left(a^2+a+1\right):\left(a^2-a-1\right) =\\ \\ = a^2+a+1\\ \\ \\ \text{Valorile admisibile ale variabilei a sunt:}\\ a\in \mathbb{R}_+^*\,\text{ si }\,a^2-a-1 \neq 0.\\\\\Delta = 1+4 = 5 \Rightarrow a\neq \dfrac{1\pm \sqrt{5}}{2}$

$\Rightarrow a \in \mathbb{R}_+^*\,\backslash\,\left\{\dfrac{1+ \sqrt{5}}{2}\right\}\quad \text{(valorile admisibile)}$