دانلود اپلیکیشن
1 حاصل عبارت های زیر را به دست آورید.
\(\begin{array}{l} - {3^2} = \\\\{\left( { - 5} \right)^2} = \\\\ - {1^5} = \\\\{\left( { - 1} \right)^5} = \\\\{\left( { - 1} \right)^4} = \\\\ - {1^4} = \\\\{7^0} = \\\\{\left( {\frac{2}{7}} \right)^0} = \\\\{\left( { - 9} \right)^0} = \\\\{0^5} + 1/{1^0} = \\\\4 + {2^0} = \\\\ - {1^5} = \\\\{2^3} = \\\\{\left( { - 2} \right)^3} = \\\\{5^0} = \end{array}\)
\( - {3^2} = - (3 \times 3) = - 9\)
\({\left( { - 5} \right)^2} = \left( { - 5} \right) \times \left( { - 5} \right) = 25\)
\( - {1^5} = - {(1)^5} = - 1\)
\(\begin{array}{l}{\left( { - 1} \right)^5} = \\\\\left( { - 1} \right) \times \left( { - 1} \right) \times \left( { - 1} \right) \times \left( { - 1} \right) \times \left( { - 1} \right) = - 1\end{array}\)
\({\left( { - 1} \right)^4} = \left( { - 1} \right) \times \left( { - 1} \right) \times \left( { - 1} \right) \times \left( { - 1} \right) = 1\)
\( - {1^4} = - {(1)^4} = - 1\)
\({7^0} = 1\)
\({\left( {\frac{2}{7}} \right)^0} = 1\)
\({\left( { - 9} \right)^0} = 1\)
\({0^5} + 1/{1^0} = 0 + 1 = 1\)
\(4 + {2^0} = 4 + 1 = 5\)
\( - {1^5} = - {(1)^5} = - 1\)
\({2^3} = 2 \times 2 \times 2 = 8\)
\({\left( { - 2} \right)^3} = \left( { - 2} \right) \times \left( { - 2} \right) \times \left( { - 2} \right) = - 8\)
\({5^0} = 1\)
2 در جای خالی علامت > یا = یا < بگذارید.
\(\begin{array}{l}{2^0}\;\bigcirc \;{2^1}\\\\{7^0}\;\bigcirc \;{7^1}\\\\{\left( {\frac{3}{5}} \right)^2}\;\bigcirc \;{2^0}\\\\\left( { - 2} \right)\;\bigcirc \;{\left( { - 2} \right)^1}\end{array}\)
\(\begin{array}{l}{2^0}\;\bigcirc \;{2^1}\\\\\left\{ \begin{array}{l}{2^0} = 1\\\\{2^1} = 2\end{array} \right. \Rightarrow {2^0}\; < \;{2^1}\end{array}\)
\(\begin{array}{l}{7^0}\;\bigcirc \;{7^1}\\\\\left\{ \begin{array}{l}{7^0} = 1\\\\{7^1} = 7\end{array} \right. \Rightarrow {7^0}\; < \;{7^1}\end{array}\)
\(\begin{array}{l}{\left( {\frac{3}{5}} \right)^2}\;\bigcirc \;{2^0}\\\\\left\{ \begin{array}{l}{\left( {\frac{3}{5}} \right)^2} = \frac{9}{{25}}\\\\{2^0} = 1\end{array} \right. \Rightarrow {\left( {\frac{3}{5}} \right)^2}\; < \;{2^0}\end{array}\)
\(\left( { - 2} \right)\;\bigcirc \;{\left( { - 2} \right)^1} \Rightarrow \left( { - 2} \right)\; = \;{\left( { - 2} \right)^1}\)
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