720 : [41-(2x-5)] =2^3 . 5

<=> 41-(2x-5)=720:40=18

<=> 2x-5=41-18=23

<=>2x=23+5=28

<=> x=28:2=14

2^x-15=17

<=> 2^x=17+15=32

<=> 2^x=2^5

<=> x=5

(7x-11)^3= 2^5.5^2+200

<=> (7x-11)^3=1000

<=> (7x-11)^3=10^3

<=> 7x-11=10

<=> 7x=10+11=21

<=> x=21:7=3

2^x-15=17 nhe. Minh viet nham

e) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3\cdot\left(2x-15\right)^2-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3\cdot\left[\left(2x-15\right)^2-1\right]=0\)

\(\Rightarrow\left(2x-15\right)^3=0\) hoặc \(\left(2x-15\right)^2-1=0\)

+)TH1: \(\left(2x-15\right)^3=0\)

\(\Rightarrow2x-15=0\)

\(\Rightarrow2x=15\)

\(\Rightarrow x=\frac{15}{2}\)

+)TH2: \(\left(2x-15\right)^2-1=0\)

\(\Rightarrow\left(2x-15\right)^2=1\)

\(\Rightarrow\left[{}\begin{matrix}2x-15=1\\2x-15=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2x=16\\2x=14\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

Vậy \(x=\frac{15}{2}\) hoặc \(x=8\) hoặc \(x=7\)

a) \(2^x-17=15\Rightarrow2^x=32\)

Mà \(2^5=32\Rightarrow x=5\)

Vậy x = 5

b)\(\left(7x-11\right)^3=2^5\cdot5^2+200\)

\(\Rightarrow\left(7x-11\right)^3=1000\)

\(\Rightarrow\left(7x-11\right)^3=10^3\)

\(\Rightarrow7x-11=10\)

\(\Rightarrow7x=21\)

\(\Rightarrow x=3\)

Vậy x = 3

c)\(x^{10}=1^x\Rightarrow x^{10}=1\)(số 1 có luỹ thừa là bao nhiêu thì vẫn là 1 thui)\(\Rightarrow x=1\)

Vậy x = 1

d) \(x^{10}=x\Rightarrow x^{10}-x=0\)

\(\Rightarrow x\left(x^9-1\right)=0\)

\(\Rightarrow x=0\) hoặc \(x^9-1=0\)

+)TH1: \(x=0\)

+)TH2: \(x^9-1=0\Rightarrow x^9=1\Rightarrow x=1\)

Vậy x = 0 hoặc x = 1

tìm x:

a)

2^x= 2⁵

=> x = 5

b)

(7x−11)³= 32 . 25 + 200

(7x−11)³= 800+200

(7x−11)³= 1000

10³= 1000

=> (7x−11)^{3 =} 10³

7x-11 = 10

7x= 11+10

7x = 21

x = 21: 7 = 3

=> x = 3

c)

=> x = 1

Vì cơ số 1 với số mũ nào thì cũng bằng 1

d)

a) 2^x-15= 17

    2^x      = 17-15

   2^x        =  2

   2^x         =  2¹

  x             =   1                      

a)

 \(\left(2x-15\right)^5=\left(2x-15\right)^3\\ \Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\\ \Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15-1\right).\left(2d-15+1\right)=0\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right. \)

b) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)

a) 2^x - 15 = 17

2^x = 2⁵

b) ( 7x - 11 )³ = 2⁵. 5² + 200

(7x - 11)³ = 32 . 25 + 200

(7x -11)³ = 1000

(7x-11)³ = 10³

7x - 11 = 10

7x = 10+11

7x = 21

x = 21 : 7

x = 3

like nha

2\(^x\)- 15 = 17

\(\Rightarrow\)2\(^x\) =  32

\(\Rightarrow\)x  = 5

2^x-15=17

=>2^x=17+15

=2^x=32

=>2^x=2^5

=>x=5

25/7nha

thì cứ tính bên phải trước rồi tìm x như thường thôi