Đáp án cần chọn là: A

Ý A nhé bạn

chúc học tốt

ko bít nữa

202³⁰³ = (202³)¹⁰¹ = 8242408¹⁰¹

303²⁰² = (303²)¹⁰¹ = 91809¹⁰¹

Do 8242408 > 91809 nên 8282408¹⁰¹ > 91809¹⁰¹

Vậy 202³⁰³ > 303²⁰²

202³⁰³ & 303²⁰²

202³⁰³ = (202³)¹⁰¹ = 8242408¹⁰¹

303²⁰² = (303²)¹⁰¹ = 91809¹⁰¹

⇒ 202³⁰³ > 303²⁰²

a) \(2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}>8^{100}\)

\(\Rightarrow2^{300}< 3^{200}\)

b) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)

c) \(3^{500}=\left(3^5\right)^{100}=243^{100}\)

\(7^{300}=\left(7^3\right)^{100}=343^{100}>243^{100}\)

\(\Rightarrow3^{500}< 7^{300}\)

Giải chi tiết giúp mình ạ~

\(\left(d\right):202^{303}=\left(202^3\right)^{101}=8242408^{101}>303^{202}=\left(303^2\right)^{101}=91809^{101}\)

\(\left(e\right):107^{50}=\left(107^2\right)^{25}=11449^{25}< 73^{75}=\left(73^3\right)^{25}=389017^{25}\)

\(\dfrac{27}{86}\) < \(\dfrac{27}{81}\) = \(\dfrac{1}{3}\) = \(\dfrac{9}{27}\) < \(\dfrac{25}{27}\) 

Vậy \(\dfrac{27}{86}\) < \(\dfrac{25}{27}\) 

`a)75<77`

`=>1/75>1/77`

`=>37/75>37/77>35/77`

`b)7779>7569`

`=>1/7779<1/7569`

`=>3734/7569>3734/7779>2099/7779`

\(\sqrt{Ai}giúpmkvớiak\)

1,

Ta có:
\(\dfrac{73}{75}=1-\dfrac{2}{75}\)

\(\dfrac{77}{79}=1-\dfrac{2}{79}\)
So sánh phân số \(\dfrac{2}{75}\) và \(\dfrac{2}{79}\)
Vì \(75< 79\) nên \(\dfrac{1}{75}>\dfrac{1}{79}\)
Vậy \(1-\dfrac{2}{75}< 1-\dfrac{2}{79}\)
Hay \(\dfrac{73}{75}< \dfrac{77}{79}\)

2,

Vì \(\dfrac{53}{100}>\dfrac{47}{100}>\dfrac{47}{106}\) nên \(\dfrac{53}{100}>\dfrac{47}{106}\)

3,

Ta có:
\(\dfrac{81}{79}=1+\dfrac{2}{79}\)

\(\dfrac{65}{63}=1+\dfrac{2}{63}\)
So sánh phân số \(\dfrac{2}{79}\) và \(\dfrac{2}{63}\)
Vì \(79>63\) nên \(\dfrac{81}{79}< \dfrac{65}{63}\)
Hay \(\Rightarrow1+\dfrac{2}{79}< 1+\dfrac{2}{63}\)
Vậy \(\dfrac{81}{79}< \dfrac{65}{63}\)

4,

\(\dfrac{48}{47}>1>\dfrac{84}{85}\)

Vậy \(\dfrac{48}{47}>\dfrac{84}{85}\)

giúp mình với ạ

cảm ơn ạ

\(\dfrac{47}{95}\) và \(\dfrac{35}{69}\)

\(\dfrac{47}{95}< \dfrac{1}{2}\) và \(\dfrac{35}{69}>\dfrac{1}{2}\)

Vậy \(\dfrac{47}{95}< \dfrac{35}{69}\)

\(\dfrac{53}{103}\) và \(\dfrac{71}{145}\)

\(\dfrac{53}{103}>\dfrac{1}{2}\) và \(\dfrac{71}{145}< \dfrac{1}{2}\)

Vậy \(\dfrac{53}{103}>\dfrac{71}{145}\)

\(\dfrac{2009}{2010}\) và \(\dfrac{2005}{2006}\)

\(1-\dfrac{2009}{2010}=\dfrac{1}{2010}\) và \(1-\dfrac{2005}{2006}=\dfrac{1}{2006}\)

Vậy \(\dfrac{2009}{2010}>\dfrac{2005}{2006}\)

\(\dfrac{783}{901}\) và \(\dfrac{738}{915}\)

\(\dfrac{738}{915}< \dfrac{783}{915}< \dfrac{783}{901}\)

Vậy \(\dfrac{783}{901}>\dfrac{738}{915}\)

nhanh giúp mình với ạ

cảm ơn bạn nhiều

a) \(\dfrac{81\times49}{7\times8}=\dfrac{81\times7\times7}{7\times8}=\dfrac{81\times7}{8}=\dfrac{567}{8}\)

b) \(\dfrac{105\times69}{3\times35}=\dfrac{3\times35\times69}{3\times35}=69\)

c) \(\dfrac{77\times39}{13\times11}=\dfrac{11\times7\times13\times3}{13\times11}=7\times3=21\)

a) \(\dfrac{81x49}{7x8}=\dfrac{81x7x7}{7x8}=\dfrac{81x7}{8}=\dfrac{567}{8}=70,875\)

b) \(\dfrac{105x69}{3x35}=\dfrac{3x35x69}{3x35}=\dfrac{69}{1}=69\)

c) \(\dfrac{77x39}{13x11}=\dfrac{11x7x3x13}{13x11}=\dfrac{7x3}{1}=\dfrac{21}{1}=21\)

a: =20/90+15/90+54/90=89/90

b: =(9/13+4/13)+2/3=1+2/3=5/3