2³⁰⁰ = (2³)¹⁰⁰ = 8¹⁰⁰ và 3²⁰⁰ = (3²)¹⁰⁰ = 9¹⁰⁰ nên 2³⁰⁰ < 3²⁰⁰;

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~

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

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

`@` `\text {Ans}`

`\downarrow`

`a)`

\(3^{200}\text{ và }2^{300}\)

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

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

Vì `9 > 8 => 9^100 > 8^100`

`=> 3^200 > 2^300`

`b)`

\(27^{101}\text{ và }81^{35}\)

\(27^{101}=\left(3^3\right)^{101}=3^{303}\)

\(81^{35}=\left(3^4\right)^{35}=3^{140}\)

Vì `303 > 140 => 3^303 > 3^140`

`=> 27^101 > 81^35`

`c)`

\(2^{332}\text{ và }3^{223}\)

\(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)

\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)

Vì `9 > 8 => 9^111 > 8^111`

`=> 2^332 < 3^223.`

a: 3^200=9^100

2^300=8^100

mà 9>8

nên 3^200>2^300

b: 27^101=3^303

81^35=3^140

mà 303>140

nên 27^101>81^35

c: 2^332<2^333=8^111

3^223>3^222=9^111

mà 9>8

nên 3^223>8^111>2^332

a) \(2^x=16=2^4\Rightarrow x=4\)

b) \(x^3=27=3^3\Rightarrow x=3\)

c) \(x^{50}=x\Rightarrow x\left(x^{49}-1\right)=0\Rightarrow x=0\) hay \(x=1\)

d) \(\left(x-2\right)^2=16=4^2\Rightarrow x-2=4\) hay \(x-2=-4\)

\(\Rightarrow x=6\) hay \(x=-2\)

a) \(2^{300}=2^{3.100}=8^{100}\)

\(3^{200}=3^{2.100}=9^{100}\)

vì \(8^{100}< 9^{100}\)

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

b) \(3^{500}=3^{5.100}=243^{100}\)

\(7^{300}=7^{3.100}=343^{100}\)

vì \(243^{100}< 343^{100}\)

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

`@` `\text {Ans}`

`\downarrow`

`1,`

`a,`

`2^x = 16`

`=> 2^x = 2^4`

`=> x = 4`

Vậy, `x = 4`

`b,`

`x^3 = 27`

`=> x^3 = 3^3`

`=> x = 3`

Vậy, `x = 3`

`c,`

\(x^{50}=x\)

`=>`\(x^{50}-x=0\)

`=>`\(x\left(x^{49}-1\right)=0\)

`=>`\(\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x^{49}=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy, `x \in {0; 1}`

`d,`

`(x-2^2)=16`

`=> x - 2^2 = 16`

`=> x = 16 + 2^2`

`=> x = 20`

Vậy, `x = 20`

`2,`

`a,`

Ta có:

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

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

Vì `8 < 9 =>`\(8^{100}< 9^{100}\)

`=>`\(2^{300}< 3^{200}\)

Vậy, \(2^{300}< 3^{200}\)

`b,`

Ta có:

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

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

Vì `243 < 343 =>`\(243^{100}< 343^{100}\)

`=>`\(3^{500}< 7^{300}\)

Vậy, \(3^{500}< 7^{300}.\)

\(\frac{2017}{2300}\)lớn hơn\(\frac{2016}{3200}\)
chúc bạn học tốt !

\(\frac{2017}{2300}\)<\(\frac{2016}{2300}\)

a)

Ta có : A = 27⁵ = (3³)⁵ = 3¹⁵

            B = 243³ = (3⁵)³ = 3¹⁵

Vì 3¹⁵ = 3¹⁵ => A = B

b )

Ta có : A = 2³⁰⁰ = (2³)¹⁰⁰ = 8¹⁰⁰

            B = 3²⁰⁰ = (3²)¹⁰⁰ = 9¹⁰⁰

Vì 8¹⁰⁰ < 9¹⁰⁰ => A<B

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 \(3^{500}< 7^{300}\)