Đáp án:
Bài 1:
a/ $12$
b/ $199$
c/ $4039$
d/ $100$
Bài 2:
a/ $2^{3000} > 3^{1000}$
b/ $4^{100} > 15^{50}$
c/ $1701.1699=1700^2-1$
Giải thích các bước giải:
Bài 1:
a/ $2^{10}:2^8.3^5:3^4$
$=2^{10-8}.3^{5-4}$
$=2^2.3$
$=12$
b/ $100^2-99^2$
$=(99+1)(99+1)-99^2$
$=99^2+99+99+1-99^2$
$=199$
c/ $2020^2-2019^2$
$=(2019+1)(2019+1)-2019^2$
$=2019^2+2019+2019+1-2019^2$
$=4039$
d/ $2^{100}:2^{98}.5^{20}:5^{18}$
$=2^{100-98}.5^{20-18}$
$=2^2.5^2$
$=4.25$
$=100$
Bài 2:
a/ $2^{3000}=2^{3.1000}=(2^3)^{1000}=8^{1000}$
Vì $8^{1000} > 3^{1000}$ nên $2^{3000} > 3^{1000}$
b/ $4^{100}=4^{2.50}=(4^2)^{50}=16^{50}$
Vì $16^{50} > 15^{50}$ nên $4^{100} > 15^{50}$
c/ $1701.1699=(1700+1)(1700-1)=1700^2+1700-1700-1=1700^2-1$
Vậy $1701.1699=1700^2-1$
