Explicităm modulul :
[tex]\it |4x-5| = \begin{cases}\it -4x+5,\ pentru\ 4x-5\ \textless \ 0 \Rightarrow x\ \textless \ \dfrac{5}{4} \Rightarrow x \in \left(-\infty,\ \dfrac{5}{4}\right)
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\it \ 4x-5,\ pentru\ 4x-5\geq0 \Rightarrow x\geq\dfrac{5}{4} \Rightarrow x \in \left[\dfrac{5}{4},\ \infty,\right)\end{cases}[/tex]
Avem două cazuri:
[tex]\it\ I)\ x\in (-\infty,\ \dfrac{5}{4}) \Rightarrow\ ecua\c{\it t}ia\ devine:
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-4x+5=1-x \Rightarrow 5-1=4x-x \Rightarrow 4 = 3x \Rightarrow x = \dfrac{4}{3}
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Dar,\ \dfrac{4}{3} \not\in \left(-\infty,\ \dfrac{5}{4}\right) \Rightarrow\ ecua\c{\it t}ia\ nu\ are \ solu\c{\it t}ie\ \^{i}n\ acest\ caz.[/tex]
[tex]\it\ II)\ x\in \left[\dfrac{5}{4},\ \infty\right) \Rightarrow\ ecua\c{\it t}ia\ devine: \\\;\\ 4x-5=1-x \Rightarrow 4x+x=1+5 \Rightarrow 5x = 6 \Rightarrow x = \dfrac{6}{5} \\\;\\ Dar,\ \dfrac{6}{5} \not\in \left[ \dfrac{5}{4},\ \infty\right) \Rightarrow\ ecua\c{\it t}ia\ nu\ are \ solu\c{\it t}ie\ \^{i}n\ acest\ caz.[/tex]
Prin urmare, ecuația dată nu admite nici o soluție.
Mulțimea soluțiilor este S = ∅.
