گام به گام کاردرکلاس صفحه 107 درس 7 ریاضی هشتم (توان و جذر)

\({8^9} \div {8^5} = {8^{9 - 5}} = {8^4}\)

\(\frac{{{3^7}}}{{{3^3}}} = {3^{7 - 3}} = {3^4}\)

\({6^3} \div 6 = {6^3} \div {6^1} = {6^{3 - 1}} = {6^2}\)

\({( - \frac{3}{5})^5} \div {( - \frac{3}{5})^2} = {( - \frac{3}{5})^{5 - 2}} = {( - \frac{3}{5})^3}\)

\(\frac{{{{( - 2)}^9}}}{{{{( - 2)}^2}}} = {( - 2)^{9 - 2}} = {( - 2)^7}\)

\(\begin{array}{l}{(\frac{8}{9})^4} \div (\frac{8}{9}) = {(\frac{8}{9})^4} \div {(\frac{8}{9})^1} = \\\\{(\frac{8}{9})^{4 - 1}} = {(\frac{8}{9})^3}\end{array}\)

\({(4/5)^6} \div {(4/5)^3} = {(4/5)^{6 - 3}} = {(4/5)^3}\)

\({( - 0/3)^5} \div {( - 0/3)^2} = {( - 0/3)^{5 - 2}} = {( - 0/3)^3}\)

\(\frac{{{7^5}}}{7} = \frac{{{7^5}}}{{{7^1}}} = {7^{5 - 1}} = {7^4}\)

\(\begin{array}{l}\frac{{{2^{15}}}}{{{{({2^3})}^\bigcirc }}} = 1 \Rightarrow \frac{{{2^{15}}}}{{{{({2^3})}^5}}} = 1\\\\{3^5} \div {3^\bigcirc } = {3^4} \Rightarrow {3^5} \div {3^1} = {3^4}\\\\{14^5} \div {(\,\,\,\,\,)^2} = {14^3} \Rightarrow {14^5} \div {(14)^2} = {14^3}\\\\\frac{{{7^3}}}{{{7^5}}} = \frac{{{7^3}}}{{{7^3} \times {7^2}}} = \frac{1}{\bigcirc } \Rightarrow \frac{{{7^3}}}{{{7^5}}} = \frac{{{7^3}}}{{{7^3} \times {7^2}}} = \frac{1}{{{7^2}}}\end{array}\)

\(\begin{array}{l}{a^{12}} \div {a^5} = {a^{12 - 5}} = {a^7}\\\\{(xy)^7} \div {(xy)^4} = {(xy)^{7 - 4}} = {(xy)^3}\\\\{( - x)^7} \div {( - x)^3} = {( - x)^{7 - 3}} = {( - x)^4}\end{array}\)