Question

 

Tìm x: 
A. ( x - 5 )² = 16
B. 4^x + 33 = 7²
C. 2^x . 4 = 128
D. 5^x . 3 - 75 = 0

Answer 1

(\(x\) - 5)² = 16

(\(x-5\))² = 4²

\(\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)

Vậy \(x\) \(\in\) { 1; 9 }

B. 4^\(x\)  + 33 = 7²

   4\(^x\) + 33 = 49

   4\(^x\)         = 49 - 33

   4\(^x\)        = 16

   4\(^x\)        = 16

  4\(^x\)        = 4²

    \(x\)        = 2

c. 2\(^x\).4 = 128

   2\(^x\)    = 128 : 4

   2\(^x\)    = 32

   2\(^x\)    = 2⁵

   \(x\)      = 5

D, 5\(^x\) . 3 - 75 = 0

    5\(^x\).3      = 75

     5\(^x\)       = 75: 3

    5\(^x\)       = 25

   5\(^x\)        = 5²

     \(x\)       = 2

Answer 2

A) \(\left(x-5\right)^2=16=4^2\Rightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)

B) \(4^x+33=7^2\Rightarrow4^x=49-33\Rightarrow4^x=16=4^2\Rightarrow x=2\)

C) \(2^x.4=128\Rightarrow2^x=32=2^5\Rightarrow x=5\)

 D) \(5^x.3-75=0\Rightarrow5^x.3=75\Rightarrow5^x=25=5^2\Rightarrow x=2\)

Answer 3

\(A.\left(x-5\right)^2=16\\ \left(x-5\right)^2=4^2\\ =>x-5=4\\ x=5+4\\ x=9\\ B.4^x+33=7^2\\ 4^x+33=49\\ 4^x=49-33\\ 4^x=16\\ 4^x=4^2\\ =>x=2\\ C.2^x\cdot4=128\\ 2^x\cdot2^2=2^7\\ 2^{x+2}=2^7\\ =>x+2=7\\ x=7-2\\ x=5\\ D.5^x\cdot3-75=0\\ 5^x\cdot3=75\\ 5^x=75:3\\ 5^x=25\\ 5^x=5^2\\ =>x=2\)