Question

Scrieti ca o singura putere, folosind factorul comun: c) 3 la puterea a 9+6.3 la puterea a 8.
d)2 la puterea a 10-2 la a 9.
e)3 la puterea a 10+2.3 la a 10.
f)3.2 la puterea a 20-2.3 la 20.
g)7.3 la puterea 15+20.3 la 15.
h)3.5 la puterea a 12 a+2.5 la 12.
i)3 la puterea 20-2.3 la 19...
ps:va rog ajutati-ma, nu mai la cate stiti sau la cate puteti!!​

Answer 1

 

$\displaystyle\bf\\ c)\\3^9+6\cdot3^8=3^{8+1}+6\cdot3^8=3^8\cdot3+6\cdot3^8=3^8(3+6)=3^8\cdot3^2=\boxed{\bf3^{10}}\\\\d)\\2^{10}-2^9=2^{9+1}-2^9=2^{9}\times2-2^9=2^9(2-1)=\boxed{\bf2^9}\\\\e)\\3^{10}+2\cdot3^{10}=3^{10}(1+2)=3^{10}\cdot3=\boxed{\bf3^{11}}\\\\f)\\3\cdot2^{20}-2\cdot2^{20}=2^{20}(3-2)=2^{20}\cdot1=\boxed{\bf2^{20}}$

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$\displaystyle\bf\\g)\\7\cdot3^{15}+20\cdot3^{15}=3^{15}(7+20)=3^{15}\cdot27=3^{15}\cdot3^3=3^{15+3}=\boxed{\bf3^{18}}\\\\h)\\3\cdot5^{12}+2\cdot5^{12}=5^{12}(3+2)=5^{12}\cdot5=5^{12+1}=\boxed{\bf5^{13}}\\\\i)\\3^{20}-2\cdot3^{19}=3^{19+1}-2\cdot3^{19}=3^{19}\cdot3-2\cdot3^{19}=3^{19}(3-2) =\boxed{\bf3^{19}}$