Question

rezolvati acel incercuit va rog

Answer 1

$x^2-3x+4=x^2-2 \cdot \frac{3}{2}x+ \frac{9}{4} + \frac{7}{4} =(x- \frac{3}{2})^2+ \frac{7}{4}\ \textgreater \ 0, \\ \\ deci~radicalul~este~de.fi.nit~pentru~orice~x \in R. \\ \\ Analizam~doua~cazuri: \\ \\ i)~x \leq 2.~In~acest~caz~x-2 \leq 0,~iar~cum~ \sqrt{x^2-3x+4}\ \textgreater \ 0, \\ \\ rezulta~ca~x \in (- \infty;2]~este~solutie.$
$ii)~x\ \textgreater \ 2.~In~acest~caz,~ambii~membri~ai~inecuatiei~sunt~ \\ \\ pozitivi~deci~putem~ridica~inecuatia~la~patrat,~obtinand: \\ \\ x^2-3x+4\ \textgreater \ x^2-4x+4 \Leftrightarrow x^2-3x+4-x^2+4x-4\ \textgreater \ 0 \Leftrightarrow \\ \\ \Leftirhgtarrow x\ \textgreater \ 0,~ceea~ce~este~adevarat,~avand~in~vedere~ca~am~presupus \\ \\ x\ \textgreater \ 2. \\ \\ Deci~x \in (2;+ \infty)~este~de~asemenea~o~solutie. \\ \\ In~concluze,~solutia~inecuatiei~este~(-\infty;2] \cup (2;+\infty)=R .$

Answer 2

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