64 . 4^x = 168

<=> 43. 4^x = 416

=> 3 + x = 16

<=> x = 13

Vậy x = 13

2^x.16² = 1024

<=> 2^x. 2⁸ = 210

=> x + 8 = 10

<=> x = 2

Vậy x = 2

b: Ta có: \(2^x\cdot16^2=1024\)

\(\Leftrightarrow2^x\cdot2^8=2^{10}\)

\(\Leftrightarrow x+8=10\)

hay x=2

  64.4^x=16³

  2⁶. (2²)^x = (2⁴)³

=> 2⁶. 2^2x = 2¹²

        2^2x   = 2 ¹² : 2⁶

       2^2x   = 2⁶

  =>  2x = 6

         x = 6 :2

         x=3

- Vậy x =3

Theo đề, ta có:

\(\dfrac{16+x}{5+x}=2\)

\(\Leftrightarrow x+16=2x+10\)

hay x=6

Vậy: \(\dfrac{x}{2}+\dfrac{1}{x-1}=\dfrac{6}{2}+\dfrac{1}{6-1}=3+\dfrac{1}{5}=\dfrac{16}{5}\)

gfjffshdunvghfuf

\(A=x^4+2x^3-16x^2-2x+15\)

\(=\left(x^4-x^2\right)+\left(2x^3-2x\right)-\left(15x^2-15\right)\)

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

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

\(=\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(x+5\right)\)

Vì x là số tự nhiên lẻ => x = 2k+1 (k thuộc N)

=>\(A=\left(2k+1-1\right)\left(2k+1+1\right)\left(2k+1-3\right)\left(2k+1+5\right)\)

\(=2k\left(2k+2\right)\left(2k-2\right)\left(2k+6\right)\)

\(=16k\left(k+1\right)\left(k-1\right)\left(k+3\right)⋮16\) (đpcm)

cung choi bang bang ak

MAU LÊN