a/ x = 4

a) 2^x.(1 + 2³) = 144

2^x . 9 = 144

2^x = 16

=> x = 4

b) (2x - 1)¹⁰ = (2x - 1)¹⁰⁰

(2x - 1)¹⁰⁰ - (2x - 1)¹⁰  = 0

 (2x - 1)¹⁰.[ (2x - 1)⁹⁰ - 1] = 0

=>  (2x - 1)¹⁰ = 0 hoặc  (2x - 1)⁹⁰ - 1 = 0

=> 2x = 1   hoặc    (2x - 1)⁹⁰ = 1

=> x = \(\frac{1}{2}\)  hoặc   \(2x-1=\orbr{\begin{cases}1\\-1\end{cases}}\)

=>                             \(2x=\orbr{\begin{cases}2\\0\end{cases}}\)

=> x = {\(\frac{1}{2};1;0\)}

a: =>2x-1=4 hoặc 2x-1=-4

=>2x=5 hoặc 2x=-3

=>x=5/2 hoặc x=-3/2

d: =>x=|2|=2

e: \(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x-y=0\end{matrix}\right.\Rightarrow x=y=1\)

\(\Leftrightarrow2x\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)

Đặt \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100};B=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)

\(\Leftrightarrow A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \Leftrightarrow A=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ \Leftrightarrow A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ \Leftrightarrow A=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}-1-\dfrac{1}{2}-\dfrac{1}{3}-...-\dfrac{1}{50}\\ \Leftrightarrow A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=B\)

\(\Leftrightarrow2x.A=B\Leftrightarrow2x.B-B=0\\ \Leftrightarrow B\left(2x-1\right)=0\\ \Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)

(2x - 1) x 10 = (2x - 1) x 100
=> 2x - 1 = 0
=> x = 1/2

\(\Rightarrow\left(2x-1\right)^{100}-\left(2x-1\right)^{80}=0\)

Đặt 2x - 1  = t ta có :

             \(t^{100}-t^{80}=0\Rightarrow t^{80}\left(t^{20}-1\right)=0\)

=> \(t^{80}=0\)  hoặc \(t^{20}-1=0\)

=> t = 0 hoặc \(t^{20}=1=\left(-1\right)^2\)

=> t= 0 hoặc t = 1 hoặc t = -1 

(+) t = 0 => 2x - 1 = 0 => x = 1/2

(+) t = 1 => 2x -  1 = 1 => x  = 1 

(+) t=  - 1 => 2x - 1 = -1 => x = 0 

tick đúng nha 

=> (2x -1)¹⁰⁰ - (2x -1)⁸⁰ = 0 

=> (2x -1)⁸⁰. [(2x-1)²⁰ - 1] = 0 

=> (2x -1)⁸⁰ = 0 hoặc (2x -1)²⁰   = 1

+)  (2x -1)⁸⁰ = 0 => 2x - 1 = 0 => x = 1/2

+)  (2x -1)²⁰   = 1 => 2x - 1 = 1 hoặc 2x - 1 = -1

=> 2x = 2 hoặc 2x = 0 => x = 1 hoặc x = 0 

Vậy....

dấu [] là giá trị tuyệt đối nha