Question

giúp mình 3 câunày với ạ . tuần sau mình cần nộp 

a) 2\(^2\).16 ≥ \(2^x\) ≥ \(4^2\)  

b) 9.27 ≤ \(3^x\) ≤ 243 

c) 2. \(\left(x-\dfrac{1}{2}\right)^2\) \(-\dfrac{1}{8}\) = 0

Answer 1

`@` `\text {Ans}`

`\downarrow`

`a)`

`2^2 * 16 \ge 2^x \ge 4^2`

`=> 2^2 * 2^4 \ge 2^x \ge 2^4`

`=> 2^6 \ge 2^x \ge 2^4`

`=> x \in {4; 5; 6}`

`b)`

`9*27 \le 3^x \le 243`

`=> 3^2 * 3^3 \le 3^x \le 3^5`

`=> 3^5 \le 3^x \le 3^5`

`=> x = 5`

`c)`

`2 * (x - 1/2)^2 - 1/8 = 0`

`=> 2* (x - 1/2)^2 = 1/8`

`=> (x - 1/2)^2 = 1/8 \div 2`

`=> (x-1/2)^2 = 1/16`

`=> (x - 1/2)^2 = (+- 1/4)^2`

`=>`\(\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{1}{4}\\x-\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{4}+\dfrac{1}{2}\\x=\dfrac{1}{2}-\dfrac{1}{4}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy, `x \in {1/4; 3/4}.`