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, \(3^{500}< 7^{300}.\)

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

c: \(3=\sqrt{16}-1< \sqrt{17}-1\)

Đáp án cần chọn là: B

a: \(11^{14}< 11^{15}\)

b: \(4^{300}=64^{100}\)

\(3^{400}=81^{100}\)

mà 64<81

nên \(4^{300}< 3^{400}\)

a, Có 3 = (3) = 9 và 2 = (2) = 8 => 3 > 2

b, Có 6 = (6) = 36 => 6 > 12

37,37 x 5959,59 = 37 x 1,01 x 59 x 101,01 = 37 x 59 x 1,01 x 101,01

59,59 x 3737,37 = 59 x 1,01 x 37 x 101,01 = 37 x 59 x 1,01 x 101,01

=> 37,37 x 5959,59 = 59,59 x 3737,37
                                              Kick mik nha

\(\dfrac{11}{-13}=-\dfrac{11}{13}=-\dfrac{13}{13}+\dfrac{2}{13}=-1+\dfrac{2}{13}\\ -\dfrac{14}{15}=-\dfrac{15}{15}+\dfrac{1}{15}=-1+\dfrac{1}{15}\)

Ta thấy : \(\dfrac{1}{15}< \dfrac{1}{13}< \dfrac{2}{13}=>-1+\dfrac{1}{15}< -1+\dfrac{2}{13}\)

hay \(\dfrac{11}{-13}>-\dfrac{14}{15}\)

a) Ta có: 3,23 < 3,32 nên -3,23 > -3,32

b) Ta có: \( - \frac{7}{3} = \frac{{ - 28}}{{12}}; - 1,25 = \frac{{ - 125}}{{100}} = \frac{{ - 5}}{4} = \frac{{ - 15}}{{12}}\)

Vì -28 < -15 nên \(\frac{{ - 28}}{{12}} < \frac{{ - 15}}{{12}}\) hay \( - \frac{7}{3}\) < -1,25

Tớ thấy : A = 1005 x 1005 = 1010025

B  = 1000 x 1010 = 1010000

Vậy A  > B

A = 1005x1005 = (1000+5)x(1010-5) =1000x1010 x 5x1010 - 5x1000 - 25 = 1000x1010 - 25 = B - 25

Vậy B > A

Sorry, bên trên viết lộn:

A = 1005x1005 = (1000+5) x (1010-5) = 1000x1010 + 5x1010 - 5x1000 - 25 = 1000x1010 + 5x10 - 25 = 1000x1010 + 25 = B + 25

Vậy A > B

a, 16^19 > 8^25

b, 27^11 > 81^8

hãy giải thích