Solutia reala a ecuatiei:
(multumesc!)

[tex] \it \sqrt{x^2-6x+9} +2|(3-x)(2x+1)| =0 \Rightarrow
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\Rightarrow \sqrt{(x-3)^2} +2|(3-x)(2x+1)| =0 \Rightarrow \sqrt{(3-x)^2} +2|(3-x)(2x+1)| =0 \Rightarrow
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\Rightarrow |3-x| +2|(3-x)|\cdot|(2x+1)| =0 \Rightarrow |3-x| (1+2|2x+1|) =0 \Rightarrow
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\Rightarrow \begin{cases}\it |3-x| =0 \Rightarrow 3-x=0 \Rightarrow x =3
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\it 1+2|2x+1| =0 \Rightarrow 2|2x+1| =-1 \ (fals)\end{cases} [/tex]

Ecuația admite soluția unică [tex] \it x = 3. [/tex]

Rayzen Rayzen · Answer 2 · 2018-07-13T16:58:47+03:00

[tex] \underset{\geq 0}{\underbrace{\sqrt{x^2-6x+9}}} +\underset{\geq 0}{\underbrace{2\big|(3-x)(2x+1)\big|}} = 0 \\ \\ \boxed{1} \quad \sqrt{x^2-6x+9} = 0 \Rightarrow x^2-6x+9 = 0 \Rightarrow (x-3)^2 = 0 \Rightarrow \\ \Rightarrow x-3 = 0 \Rightarrow x = 3. \\ \\ \boxed{2} \quad 2\cdot |(3-x)(2x+1)| = 0 \Rightarrow (3-x)(3x+1) = 0 \Rightarrow \\ \Rightarrow (3-x = 0) ~\text{sau}~(2x+1 = 0) \Rightarrow \\ \Rightarrow (x = 3)~\text{sau}~(2x = -1) \Rightarrow (x = 3)~\text{sau}~\Big(x = -\dfrac{1}{2}\Big) \Rightarrow [/tex][tex]\Rightarrow x \in \Big\{-\dfrac{1}{2}, 3\Big\} \\ \\ \\ \Rightarrow S =\boxed{1}\cap \boxed{2} \Leftrightarrow \{3\} \cap\Big\{-\dfrac{1}{2},3\Big\} = \{3\} [/tex]