Giải thích các bước giải:

2/ a/ $25^{13}.125^7=(5^2)^{13}.(5^3)^7=5^{26}.5^{21}=5^{47}$

b/ $49^{48}:7^{94}=(7^2)^{48}:7^94=7^{96}:7^{94}=7^2$

c/ $(\dfrac{8}{27})^{41}.(\dfrac{4}{9})^{69}$

$=[(\dfrac{2}{3})^3]^{41}.[(\dfrac{2}{3})^2]^{69}$

$=(\dfrac{2}{3})^{123}.(\dfrac{2}{3})^{138}$

$=(\dfrac{2}{3})^{261}$

d/ $(\dfrac{25}{16})^{71}:(\dfrac{125}{64})^{20}$

$=[(\dfrac{5}{4})^2]^{71}:[(\dfrac{5}{4})^3]^{20}$

$=(\dfrac{5}{4})^{142}:(\dfrac{5}{4})^{60}$

$=(\dfrac{5}{4})^{82}$

3/ a/ $2^{500}=(2^5)^{100}=32^{100}$

và $5^{200}=(5^2)^{100}=25^{100}$

Vì $32^{100}>25^{100}$ nên $2^{500}>5^{200}$

b/ $2^{3333}=(2^3)^{1111}=8^{1111}$

và $3^{2222}=(3^2)^{1111}=9^{1111}$

Vì $8^{1111}<9^{1111}$ nên $2^{3333}<3^{2222}$

c/ $4^{2000}=(4^2)^{1000}=16^{1000}$

và $2^{4000}=(2^4)^{1000}=16^{1000}$

Vì $16^{1000}=16^{1000}$ nên $4^{2000}=2^{4000}$

d/ $99^{20}=(99^2)^{10}=9801^{20}$

Vì $9801^{20}<9999^{10}$ nên $99^{20}<9999^{10}$

Chúc bạn học tốt !!!

Đáp án:

Giải thích các bước giải:

$25^{13}.125^7=5^{26}.5^{21}=5^{47}$

$49^{48}:7^{94}=7^{96}:7^{94}=7^2$

$\left (\dfrac{8}{27} \right )^{41}.\left (\dfrac{4}{9} \right )^{69}$

$=\left (\dfrac{2}{3} \right )^{123}.\left (\dfrac{2}{3} \right )^{138}$

$=\left (\dfrac{2}{3} \right )^{261}$

$\left (\dfrac{25}{16} \right )^{71}:\left (\dfrac{125}{64} \right )^{20}$

$=\left (\dfrac{5}{4} \right )^{142}:\left (\dfrac{5}{4} \right )^{60}$

$=\left (\dfrac{5}{4} \right )^{82}$

$3.$

$a,2^{500}=32^{100}$

$5^{200}=25^{100}$

$⇒2^{500}>5^{200}$

$b,2^{3333}=8^{1111}$

$3^{2222}=9^{1111}$

$⇒2^{3333}<3^{2222}$

$c,4^{2000}=16^{1000}$

$2^{4000}=16^{1000}$

$⇒4^{2000}=2^{2000}$

$d,99^{20}=(99^2)^{10}=9801^{10}$

$⇒99^{20}<9999^{10}$