Question

Решите пожалуйста, номера 4.7 и 3.18) Срочно!))

Answer 1

$\frac{(-5)^6*(-5)^7*(-5)^8}{(-5)^{14}*(-5)^4}= \frac{(-5)^{21}}{(-5)^{18}} =(-5)^{21-18}=(-5)^3=-125$

$\frac{1,2^{40}*1,2^{25}*1,2^4}{1,2^{59}*1,2^8} =1,2^{40+25+4-59-8}=1,2^2=1,44$

$\frac{ (\frac{1}{3} )^{10}* (\frac{1}{3} )^{20}* (\frac{1}{3} )^{30}}{ (\frac{1}{3} )^{34}* (\frac{1}{3} )^{23}}= (\frac{1}{3} )^{10+20+30-34-23}= (\frac{1}{3} )^3= \frac{1}{27}$

$\frac{ (-\frac{1}{6} )^{25}* (-\frac{1}{6} )^{19}* (-\frac{1}{6} )^{16}}{ (-\frac{1}{6} )^{8}* (-\frac{1}{6} )^{49}}= (-\frac{1}{3} )^{25+19+16-8-49}= (-\frac{1}{6} )^3= -\frac{1}{216}$

(5²)²-600=5⁴-600=625-600=0
(3³)²-271=3⁶-271=729-271=458
1000-5*(2³)²=1000-5*2⁶=1000-5*2*2⁵)=1000-10*32=10(100-32)=10*68=680
$( (\frac{1}{2})^4)^2*320=2^{-8}*32*10=2^{-8}*2^5*2*5=2{-2}*5= \frac{5}{4} =1,25$
(2⁴)²-200=2⁸-200=256-200=56
$(( \frac{2}{3})^2)^3 * \frac{3645}{32} =\frac{2^6}{3^6} * \frac{5*729}{2^5} =\frac{2}{3^6} * 5*3^6 =10$