Question

Tĩm x:

a) 2(2x-1)=x-2

b) (3x-5)^3= (3x-5)^2

c) |3x-5|+ |3x+1|=6 với x nguyên

Answer 1

a; 2.(2\(x\) - 1) = \(x-2\)

     4\(x\) - 2 = \(x\) - 2

     4\(x\) - \(x\) = -2 + 2

      3\(x\)= 0

        \(x\) = 0

Vậy \(x\) = 0

b; (3\(x\) - 5)³ = (3\(x\) - 5)²

    (3\(x\) - 5)³ - (3\(x\) - 5)² = 0

   (3\(x-5\))².(3\(x\) - 5 - 1) = 0

     \(\left[{}\begin{matrix}3x-5=0\\3x-5-1=0\end{matrix}\right.\)

      \(\left[{}\begin{matrix}3x=5\\3x=6\end{matrix}\right.\)

       \(\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{6}{3}\end{matrix}\right.\)

        \(\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=2\end{matrix}\right.\)

Vậy \(x\) {\(\dfrac{5}{3}\); \(2\)}