вправа 1.12 гдз 11 клас математика Мерзляк Номіровський 2019
Вправа 1.12
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\begin{equation} 1) (a^{\sqrt{5}}+2)(a^{\sqrt{5}}-2)-(a^{\sqrt{3}}+3)^{2}= \end{equation} \begin{equation} =(a^{\sqrt{5}})^{2}-2^{2}-(a^{\sqrt{5}})^{2}-6a^{\sqrt{5}}-9= \end{equation} \begin{equation} =-6a^{\sqrt{5}}-13; \end{equation} \begin{equation} 2) \frac{a^{2\sqrt{3}}-b^{2\sqrt{2}}}{(a^{\sqrt{3}}+b^{\sqrt{2}})^{2}}+1= \end{equation} \begin{equation} =\frac{(a^{\sqrt{3}^{2}}-(b^{\sqrt{2}})^{2}}{(a^{\sqrt{3}}+b^{\sqrt{2}})^{2}}+1= \end{equation} \begin{equation} =\frac{(a^{\sqrt{3}}+b^{\sqrt{2}})(a^{\sqrt{3}}-b^{\sqrt{2}})}{(a^{\sqrt{3}}+b^{\sqrt{2}})^{2}}+1= \end{equation} \begin{equation} =\frac{a^{\sqrt{3}}-b^{\sqrt{2}}}{a^{\sqrt{3}}+b^{\sqrt{2}}}= \end{equation} \begin{equation} =\frac{a^{\sqrt{3}}-b^{\sqrt{2}}}{a^{\sqrt{3}}+b^{\sqrt{2}}}+1= \end{equation} \begin{equation} =\frac{a^{\sqrt{3}}-b^{\sqrt{2}}+a^{\sqrt{3}}+b^{\sqrt{2}}}{a^{\sqrt{3}}+b^{\sqrt{2}}}= \end{equation} \begin{equation} =\frac{2a^{\sqrt{3}}}{a^{\sqrt{3}}+b^{\sqrt{2}}}. \end{equation}
\begin{equation} 1) (a^{\sqrt{5}}+2)(a^{\sqrt{5}}-2)-(a^{\sqrt{3}}+3)^{2}= \end{equation} \begin{equation} =(a^{\sqrt{5}})^{2}-2^{2}-(a^{\sqrt{5}})^{2}-6a^{\sqrt{5}}-9= \end{equation} \begin{equation} =-6a^{\sqrt{5}}-13; \end{equation} \begin{equation} 2) \frac{a^{2\sqrt{3}}-b^{2\sqrt{2}}}{(a^{\sqrt{3}}+b^{\sqrt{2}})^{2}}+1= \end{equation} \begin{equation} =\frac{(a^{\sqrt{3}^{2}}-(b^{\sqrt{2}})^{2}}{(a^{\sqrt{3}}+b^{\sqrt{2}})^{2}}+1= \end{equation} \begin{equation} =\frac{(a^{\sqrt{3}}+b^{\sqrt{2}})(a^{\sqrt{3}}-b^{\sqrt{2}})}{(a^{\sqrt{3}}+b^{\sqrt{2}})^{2}}+1= \end{equation} \begin{equation} =\frac{a^{\sqrt{3}}-b^{\sqrt{2}}}{a^{\sqrt{3}}+b^{\sqrt{2}}}= \end{equation} \begin{equation} =\frac{a^{\sqrt{3}}-b^{\sqrt{2}}}{a^{\sqrt{3}}+b^{\sqrt{2}}}+1= \end{equation} \begin{equation} =\frac{a^{\sqrt{3}}-b^{\sqrt{2}}+a^{\sqrt{3}}+b^{\sqrt{2}}}{a^{\sqrt{3}}+b^{\sqrt{2}}}= \end{equation} \begin{equation} =\frac{2a^{\sqrt{3}}}{a^{\sqrt{3}}+b^{\sqrt{2}}}. \end{equation}