номер 247 гдз 8 класс алгебра Муравин Муравин

1)\begin{equation}\sqrt{a^{2}}+\sqrt{b^{2}}  \sqrt{a^{2}+b^{2}}  \sqrt{(a+b)^{2}} \end{equation}
a = 3  b = 4 \begin{equation}\sqrt{a^{2}}+\sqrt{b^{2}}=\sqrt{3^{2}}+\sqrt{4^{2}}=3+4=7 \end{equation} \begin{equation}\sqrt{a^{2}+b^{2}}=\sqrt{3^{2}+4^{2}}= \end{equation} \begin{equation}=\sqrt{9+16}=\sqrt{25}=5 \end{equation} \begin{equation}\sqrt{\left ( a+b \right )^{2}}=\sqrt{(3+4)^{2}}=\sqrt{7^{2}}=7 \end{equation}
2)\begin{equation}\sqrt{a}+\sqrt{b}  2\sqrt{ab}  \sqrt{a+b}  \end{equation}
a = 16 b = 9 \begin{equation}\sqrt{a}+\sqrt{b}=\sqrt{16}+\sqrt{9}=4+3=7 \end{equation} \begin{equation}2\sqrt{ab}=2\cdot \sqrt{16\cdot 9}=2\cdot 4\cdot 3=24 \end{equation} \begin{equation}\sqrt{a+b}=\sqrt{16+9}=\sqrt{25}=5 \end{equation}
Ответ:
1) \begin{equation}\sqrt{a^{2}}+\sqrt{b^{2}}=\sqrt{(a+b)^{2}}> \sqrt{a^{2}+b^{2}} \end{equation}
2)\begin{equation}2\sqrt{ab}> \sqrt{a}+\sqrt{b}> \sqrt{a+b} \end{equation}