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Rezolvati exercitiul 10​

Rezolvati Exercitiul 10 class=

Răspuns :

Accesează poza pt rezolvare. Succes!

Vezi imaginea BIBISOFI

a)

[tex]\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\}[/tex]

b)

[tex]\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\}[/tex]

c)

[tex]\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)[/tex]

d)

[tex]\it |x-8|=0 \Rightarrow x-8=0 \Rightarrow x=8[/tex]

e)

[tex]\it |x+18|=0 \Rightarrow x+18=0 \Rightarrow x=-18[/tex]

f)

[tex]\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[/tex]