Question

|x^2-36|+|6-x|=0 cum calculez

Answer 1

$|x^2-36|+|6-x|=0 \Rightarrow \big|(x-6)(x+6)\big| + |6-x| = 0\Rightarrow \\ \\ Conform proprietatii: \quad |a\cdot b| = |a| \cdot |b| \\ \\ \Rightarrow |x-6|\cdot|x+6|+|6-x| = 0 \\ \\ Conform proprietatii: \quad |a| = |-a| \\ \\ \Rightarrow |x-6|\cdot |x+6| +\big|-(6-x)\big| = 0 \Rightarrow |x-6|\cdot|x+6|+|x-6|=0\Rightarrow \\ \RIghtarrow |x-6|\cdot\big(|x+6|+1)=0 \\ \\ \boxed{1} \quad |x-6| = 0 \Rightarrow x-6 = \pm0 \Rightarrow x-6 =0 \Rightarrow \boxed{x_1=6}$

$\boxed{2} \quad |x+6|+1 = 0 \Rightarrow |x+6| = -1 \quad (F) \Rightarrow x\in \O \\ \\ \text{Fals deoarece modulul unui numar nu poate sa dea un numar negativ} \\ \\ \\ Din \boxed{1} \cup \boxed{2} \Rightarrow S=\{6\}$