ai giúp mình nhanh đc ko

\(a,3^{39}=\left(3^3\right)^{13}=9^{13}< 11^{13}< 11^{21}\\ b,2^{91}=\left(2^{13}\right)^7=8192^7< 3125^7=\left(5^5\right)^7=5^{35}\)

Đây bạn nhé! Chữ hơi xấu, thông cảm

(x-1)²⁰²⁰=(x-1)²⁰²²

=>(x-1)²⁰²⁰-(x-1)²⁰²²=0

=>(x-1)²⁰²⁰-(x-1)²⁰²⁰.(x-1)²=0

=>(x-1)²⁰²⁰(1-(x-1)²=0

=>(x-1)²⁰²⁰=0 hoặc 1-(x-1)²=0

=>x=1 hoặc x=2.

Bài 2

a,2¹⁰⁵ và 5⁴⁵

2¹⁰⁵=(2⁷)¹⁵=128¹⁵

5⁴⁵=(5³)¹⁵=125¹⁵

Vì 128^15_>12515 nên 2¹⁰⁵>5⁴⁵.

b,

5⁵⁴ và 3⁸¹

5⁵⁴=(5⁶)⁹=15625⁹

3⁸¹=(3⁹)⁹=19683⁹

Vì 15625⁹<19683⁹ nên 5⁵⁴<3⁸¹

Bài 1 :

\(\left(x-1\right)^{2020}=\left(x-1\right)^{2022}\)

\(\Rightarrow\left(x-1\right)^{2022}-\left(x-1\right)^{2020}=0\)

\(\Rightarrow\left(x-1\right)^{2020}\left[\left(x-1\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=1\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

2⁹¹ và 5³⁵

2⁹¹ = (2¹³)⁷ = 8192⁷

5³⁵ = (5⁵)⁷ = 3125⁷

Vì 8192⁷ > 3125⁷ => 2⁹¹ > 5³⁵

Vậy 2⁹¹ > 5³⁵

2⁹¹ <  5³⁵

2 mũ 21>5 mũ35

Ta có: 291 > 290 = (25)18 = 3218

535 < 536 = (52)18 = 2518.

Vì 32 > 25 nên 3218 > 2518, do đó ta có : 291 > 3218 > 2518 > 535.

Vậy 291 > 535.

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

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

mà -1/32>-1/27

nên a>b

a>b

291 < 535

291 < 535 

291 < 535

`2^{91}=(2^{13})^{7}=8192^{7}`

`5^{35}=(5^{5})^{7}=3125^{7}`

Vì `8192^{7}>3125^{7}`

`->2^{91}>5^{35}`

\(2^{91}=\left(2^{13}\right)^7=8192^7\)

\(5^{35}=\left(5^5\right)^7=3125^7\)

Mà \(8192^7>3125^7\Rightarrow2^{91}>5^{35}\)

\(2^{91}=\left(2^{13}\right)^7\)

\(5^{35}=\left(5^5\right)^7\)

mà \(2^{13}>5^5\)

nên \(2^{91}>5^{35}\)