Question

Rezolvati exercitiul 10​

Answer 1

Accesează poza pt rezolvare. Succes!

Answer 2

a)

$\it |x-1| =\dfrac{2}{3} \Rightarrow x-1=\pm\dfrac{2}{3} \Rightarrow x-1\in\Big\{-\dfrac{2}{3},\ \dfrac{2}{3} \Big\}|_{+1} \Rightarrow x\in\Big\{\dfrac{1}{3},\ \dfrac{5}{3}\Big\}$

b)

$\it |x+5| =\dfrac{3}{4} \Rightarrow x+5=\pm\dfrac{3}{4} \Rightarrow x+5\in\Big\{-\dfrac{3}{4},\ \dfrac{3}{4} \Big\}|_{-5} \Rightarrow x\in\Big\{-\dfrac{23}{4},\ -\dfrac{17}{4}\Big\}$

c)

$\it |x-2,(3)|(0,75+|x-7|) =0\ \ \ \ \ (*)\\ \\ 0,75+|x-7|>0 \stackrel{(*)}{\Longrightarrow}\ |x-2,(3)|=0 \Rightarrow x-2,(3)=0 \Rightarrow x=2,(3)$

d)

$\it |x-8|=0 \Rightarrow x-8=0 \Rightarrow x=8$

e)

$\it |x+18|=0 \Rightarrow x+18=0 \Rightarrow x=-18$

f)

$\it |x-3|+|x^2-9| =0\Rightarrow |x-3|+|(x-3)(x+3)|=0 \Rightarrow\\ \\ \Rightarrow |x-3|+|x-3|\cdot|x+3|=0 \Rightarrow |x-3|(1+|x+3|)=0\ \ \ \ \ (*)\\ \\ 1+|x+3|>0 \stackrel{(*)}{\Longrightarrow} |x-3|=0 \Rightarrow x-3=0 \Rightarrow x=3$