вправа 2.20 гдз 11 клас математика Мерзляк Номіровський 2019
Вправа 2.20
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Відповідь ГДЗ:
\begin{equation} 1)4*9^{x}-7*12^{x}+3*16^{x}=0; \end{equation} \begin{equation} 4*3^{2x}-7*3^{x}*4^{x}+3*4^{2x}=0; \end{equation} \begin{equation} 4*(\frac{3}{4})^{2x}+7*(\frac{3}{4})^{x}+3=0; \end{equation} \begin{equation} (\frac{3}{4})^{x}=t; \end{equation} \begin{equation} 4t^{2}-7t+3=0; \end{equation} \begin{equation} D=(-7)^{2}-4*4*3= \end{equation} \begin{equation} =49-48=1. \end{equation} \begin{equation} t_1=\frac{7+1}{8}=1; \end{equation} \begin{equation} t_2=\frac{7-1}{8}= \end{equation} \begin{equation} =\frac{6}{8}=\frac{3}{4}. \end{equation} \begin{equation} \begin{bmatrix} (\frac{3}{4})^{x}=1, \\ (\frac{3}{4})^{x}=\frac{3}{4}; \end{bmatrix} \begin{bmatrix} x=0, \\ x=1. \end{bmatrix} \end{equation} \begin{equation} 2)5*2^{x}+2*5^{x}=7*10^{\frac{x}{2}}; \end{equation} \begin{equation} 5*(\sqrt{2})^{2x}-7*(\sqrt{2})^{x}+ \end{equation} \begin{equation} +2*(\sqrt{5})^{x}=0; \end{equation} \begin{equation} 5*(\frac{\sqrt{2}}{\sqrt{5}})^{2x}- \end{equation} \begin{equation} -7*(\frac{\sqrt{2}}{\sqrt{5}})^{x}+2=0; \end{equation} \begin{equation} (\frac{\sqrt{2}}{\sqrt{5}})^{x}=a. \end{equation} \begin{equation} 5a^{2}-7a+2=0; \end{equation} \begin{equation} D=(-7)^{2}-4*5*2= \end{equation} \begin{equation} 49-40=9; \end{equation} \begin{equation} a_1=\frac{7+3}{10}=1; \end{equation} \begin{equation} a_2=\frac{7-3}{10}= \end{equation} \begin{equation} =\frac{4}{10}=\frac{2}{5}. \end{equation} \begin{equation} \begin{bmatrix} (\frac{\sqrt{2}}{\sqrt{5}})^{x}=1, \\ (\frac{\sqrt{2}}{\sqrt{5}})^{x}=\frac{2}{5}; \end{bmatrix} \end{equation} \begin{equation} \begin{bmatrix} (\frac{\sqrt{2}}{\sqrt{5}})^{x}=(\frac{\sqrt{2}}{\sqrt{5}})^{0}, \\ (\frac{\sqrt{2}}{\sqrt{5}})^{x}=(\frac{\sqrt{2}}{\sqrt{2}})^{2}; \end{bmatrix} \begin{bmatrix} x=0, \\ x=2. \end{bmatrix} \end{equation}
\begin{equation} 1)4*9^{x}-7*12^{x}+3*16^{x}=0; \end{equation} \begin{equation} 4*3^{2x}-7*3^{x}*4^{x}+3*4^{2x}=0; \end{equation} \begin{equation} 4*(\frac{3}{4})^{2x}+7*(\frac{3}{4})^{x}+3=0; \end{equation} \begin{equation} (\frac{3}{4})^{x}=t; \end{equation} \begin{equation} 4t^{2}-7t+3=0; \end{equation} \begin{equation} D=(-7)^{2}-4*4*3= \end{equation} \begin{equation} =49-48=1. \end{equation} \begin{equation} t_1=\frac{7+1}{8}=1; \end{equation} \begin{equation} t_2=\frac{7-1}{8}= \end{equation} \begin{equation} =\frac{6}{8}=\frac{3}{4}. \end{equation} \begin{equation} \begin{bmatrix} (\frac{3}{4})^{x}=1, \\ (\frac{3}{4})^{x}=\frac{3}{4}; \end{bmatrix} \begin{bmatrix} x=0, \\ x=1. \end{bmatrix} \end{equation} \begin{equation} 2)5*2^{x}+2*5^{x}=7*10^{\frac{x}{2}}; \end{equation} \begin{equation} 5*(\sqrt{2})^{2x}-7*(\sqrt{2})^{x}+ \end{equation} \begin{equation} +2*(\sqrt{5})^{x}=0; \end{equation} \begin{equation} 5*(\frac{\sqrt{2}}{\sqrt{5}})^{2x}- \end{equation} \begin{equation} -7*(\frac{\sqrt{2}}{\sqrt{5}})^{x}+2=0; \end{equation} \begin{equation} (\frac{\sqrt{2}}{\sqrt{5}})^{x}=a. \end{equation} \begin{equation} 5a^{2}-7a+2=0; \end{equation} \begin{equation} D=(-7)^{2}-4*5*2= \end{equation} \begin{equation} 49-40=9; \end{equation} \begin{equation} a_1=\frac{7+3}{10}=1; \end{equation} \begin{equation} a_2=\frac{7-3}{10}= \end{equation} \begin{equation} =\frac{4}{10}=\frac{2}{5}. \end{equation} \begin{equation} \begin{bmatrix} (\frac{\sqrt{2}}{\sqrt{5}})^{x}=1, \\ (\frac{\sqrt{2}}{\sqrt{5}})^{x}=\frac{2}{5}; \end{bmatrix} \end{equation} \begin{equation} \begin{bmatrix} (\frac{\sqrt{2}}{\sqrt{5}})^{x}=(\frac{\sqrt{2}}{\sqrt{5}})^{0}, \\ (\frac{\sqrt{2}}{\sqrt{5}})^{x}=(\frac{\sqrt{2}}{\sqrt{2}})^{2}; \end{bmatrix} \begin{bmatrix} x=0, \\ x=2. \end{bmatrix} \end{equation}