номер 163 гдз 8 класс алгебра Муравин Муравин

1)\begin{equation}x^{-1}+y^{-1} \end{equation}
и \begin{equation}(x+y)^{-1} \end{equation}
при x = -2 y = 3 \begin{equation}x^{-1}+y^{-1}=\frac{1}{x}+\frac{1}{y}= \end{equation} \begin{equation}=-\frac{1}{2}+\frac{1}{3}=\frac{-3+2}{6}=-\frac{1}{6} \end{equation} \begin{equation}(x+y)^{-1}=\frac{1}{(x+y)}=\frac{1}{-2+3}=\frac{1}{1}=1 \end{equation} \begin{equation}(x+y)^{-1}>(x^{-1}+y^{-1}) \end{equation} \begin{equation}1> -\frac{1}{6} \end{equation}
2)\begin{equation}a^{-2}+b^{-2} \end{equation}
и \begin{equation}(a+b)^{-2} \end{equation}
при \begin{equation}a=\frac{1}{3}    b=-\frac{1}{5} \end{equation} \begin{equation}a^{-2}+b^{-2}=\frac{1}{a^{2}}+\frac{1}{b^{2}}= \end{equation} \begin{equation}=\frac{1}{\left ( \frac{1}{3} \right )^{2}}+\frac{1}{\left ( -\frac{1}{5} \right )^{2}}=\frac{1}{\frac{1}{9}}+\frac{1}{\frac{1}{25}}= \end{equation} \begin{equation}=9+25=34 \end{equation} \begin{equation}(a+b)^{-2}=\frac{1}{(a+b)^{2}}= \end{equation} \begin{equation}=\frac{1}{\left ( \frac{1}{3}-\frac{1}{5} \right )^{2}}=\frac{1}{\left ( \frac{5-3}{15} \right )^{2}}= \end{equation} \begin{equation}=\frac{1}{\left ( \frac{2}{15} \right )^{2}}=\frac{1}{\frac{4}{225}}= \end{equation} \begin{equation}=\frac{225}{4}=56,25 \end{equation} \begin{equation}56,25> 34 \end{equation} \begin{equation}\Rightarrow \left ( a+b \right )^{-2}> \left ( a^{-2}+b^{-2} \right ) \end{equation}
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