a, $( 2-x)³= ( 2-x)²$
⇔ $( 2-x)³-( 2-x)²= 0$
⇔ $( 2-x)².( 2-x-1)= 0$
⇔ $( 2-x)².( 1-x)= 0$
⇔ \(\left[ \begin{array}{l}( 2-x)²=0\\1-x=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=2\\x=1\end{array} \right.\)
Vậy ...
b, $( x-1)^{5}= ( 1-x)³$
⇔ $( x-1)^{5}+( x-1)³= 0$
⇔ $( x-1)³.[( x-1)²+1]= 0$
VÌ $( x-1)²+1$≥ 1> 0 ∀x
⇒ $( x-1)³= 0$
⇔ $x= 1$
c, $( 2x-1)^{7}= 3^{7}$
Vì mũ lẻ
⇒ $2x-1= 3$
⇔ $x= 2$