Ta có : \(11^{1979}< 11^{1980}=\left(11^3\right)^{660}=1331^{660}\)

            \(37^{1320}=\left(37^2\right)^{660}=1329^{660}\)

Vì \(1329^{660}>1331^{660}\) nên \(11^{1979}< 37^{1320}\)

11¹⁹⁷⁹<11¹⁹⁸⁰=1331⁶⁶⁰

37¹³²⁰=1369⁶⁶⁰

Vì 1369⁶⁶⁰>1331⁶⁶⁰ nên 37¹³²⁰>11¹⁹⁷⁹

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~

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

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

a >

B <

a)Ta có : 404303/303202=1+101101/303202

303202/202101=1+101101/202101

Do 101101/303202<101101/202101 ⇒404303/303202>303202/202101

a: \(2^{300}=8^{100}\)

\(3^{200}=9^{100}\)

mà 8<9

nên \(2^{300}< 3^{200}\)

b: \(3^{500}=243^{100}\)

\(7^{300}=343^{100}\)

mà 243<243

nên \(3^{500}< 7^{300}\)

a, Ta có : \(8>7\)

\(\Rightarrow2^{13}.8=2^{16}>2^{13}.7\)

b, Ta có : \(199^{20}< 200^{20}=2^{60}.5^{40}\)

Mà \(2003^{15}>2000^{15}=2^{60}.2^{45}\)

Thấy : \(45>40\)

\(\Rightarrow2000^{15}>200^{20}\)

\(\Rightarrow2003^{15}>199^{20}\)

c, Ta có : \(\left\{{}\begin{matrix}202^{303}=\left(2.101\right)^{3.101}=\left(8.101^3\right)^{101}\\303^{202}=\left(3.101\right)^{2.101}=\left(9.101^2\right)^{101}\end{matrix}\right.\)

Mà \(8.101^3>9.101^2\)

\(\Rightarrow202^{303}>303^{202}\)

a) Ta có: \(2^{16}=2^{13}\cdot8\)

mà \(7< 8\)

nên \(7\cdot2^{13}< 2^{16}\)

b) \(199^{20}=1568239201^5\)

\(2003^{15}=8036054027^5\)

mà \(1568239201< 8036054027\)

nên \(199^{20}< 2003^{15}\)

c) Ta có: \(202^{303}=\left(202^3\right)^{101}\)

\(303^{202}=\left(303^2\right)^{101}\)

mà \(202^3>303^2\)

nên \(202^{303}>303^{202}\)

\(\frac{72}{167}>\frac{71}{169}\)

\(\frac{63}{314}< \frac{65}{321}\)

\(\frac{25}{101}< \frac{46}{180}\)

\(a=\left[\left(-\dfrac{1}{2}\right)^5\right]^{107}=\left(-\dfrac{1}{32}\right)^{107}\)

\(b=\left[\left(-\dfrac{1}{3}\right)^3\right]^{107}=\left(-\dfrac{1}{27}\right)^{107}\)

mà -1/32>-1/27

nên a>b

a>b