گام به گام تمرین صفحه 104 درس 7 ریاضی هشتم (توان و جذر)

\( = {8^3} = {({2^3})^3} = {2^9}\) حجم مکعب با ضلع 8

\(27 \times {9^5} = {3^3} \times {({3^2})^5} = {3^3} \times {3^{10}} = {3^{13}}\)

\(\begin{array}{l}{\left[ {{{( - 3)}^2}} \right]^2} = {( - 3)^{2 \times 2}} = {( - 3)^4} = {3^4}\\\\{\left[ {{{(3)}^2}} \right]^2} = {(3)^{2 \times 2}} = {3^4}\end{array}\)

\(\begin{array}{l}{2^5} \times {2^2} \times {3^7} \times {6^3} = {2^{5 + 2}} \times {3^7} \times {6^3} = \\\\{2^7} \times {3^7} \times {6^3} = {(2 \times 3)^7} \times {6^3} = \\\\{6^7} \times {6^3} = {6^{7 + 3}} = {6^{10}}\end{array}\)

\(\begin{array}{l}{( - \frac{5}{6})^3} \times {(\frac{7}{5})^3} \times {(\frac{3}{7})^3} = \\\\{\left[ {( - \frac{5}{6}) \times (\frac{7}{5}) \times (\frac{3}{7})} \right]^3} = \\\\{\left[ {( - \frac{3}{6})} \right]^3} = {( - \frac{1}{2})^3}\end{array}\)

\({2^{40}} \times {2^{40}} \times {2^{40}} = {2^{40 + 40 + 40}} = {2^{120}}\)

\(\begin{array}{l}{({x^2})^5}.{({y^3})^2}.{x^3}{y^4} = {x^{2 \times 5}}.{y^{3 \times 2}}.{x^3}{y^4} = \\\\{x^{10 + 3}}.{y^{6 + 4}} = {x^{13}}.{y^{10}}\end{array}\)

\(16 = {( - 2)^4} > 15\)

\({2^7} + {2^7} = 2 \times {2^7} = {2^1} \times {2^7} = {2^{1 + 7}} = {2^8}\)

\(\begin{array}{l}{3^5} + {3^5} + {3^5} = 3 \times {3^5} = {3^1} \times {3^5} = \\\\{3^{1 + 5}} = {3^6}\end{array}\)

\(3 \times {(2)^3} = {5^2} - {(1)^4}\)

 الف

\(\begin{array}{l} - 2(a + {b^2}) + \frac{{{c^2}}}{b} = \\\\ - 2(6 + {( - 2)^2}) + \frac{{{{( - 4)}^2}}}{{( - 2)}} = \\\\ - 2(6 + 4) - 8 = - 28\end{array}\)

ب

\(\begin{array}{l}{(\frac{a}{b})^2} + bc - 10 = \\\\{(\frac{6}{{ - 2}})^2} + ( - 2)( - 4) - 10 = \\\\9 + 8 - 10 = 7\end{array}\)