8-|1-3x|=3
⇔ |1-3x|=5
⇔ \(\left[ \begin{array}{l}1-3x=5\\1-3x=-5\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}3x=-4\\3x=6\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-\frac{4}{3} \\x=2\end{array} \right.\)
|x-1|+3=2.x
⇔ |x-1|=2x-3
⇔ \(\left[ \begin{array}{l}x-1=2x-3\\x-1=3-2x\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}-x=-2\\3x=4\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=\frac{4}{3} \end{array} \right.\)
(0,4x-1,3)²=5,29
⇔ \(\left[ \begin{array}{l}0,4x-1,3=2,3\\0,4x-1,3=-2,3\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}0,4x=3,6\\0,4x=2,6\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=9\\x=6,5\end{array} \right.\)
(x-3)^10=(x-3)^11
⇔ (x-3)^10 - (x-3)^11 = 0
⇔ (x-3)^10.[1-(x-3)] =0
⇔ (x-3)^10.(1-x+3) = 0
⇔ (x-3)^10.(4-x)=0
⇔ \(\left[ \begin{array}{l}(x-3)^{10}=0\\4-x=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=3\\x=4\end{array} \right.\)
($\frac{1}{2}$ )²x - 1 = 1/8
⇔ $\frac{1}{4}$x = 9/8
⇔ x = $\frac{9}{2}$
