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x²+6x+10 mai mare sau egal cu 1

Răspuns :

[tex]x^{2} + 6x + 10 \geq 1 =\ \textgreater \ x^{2} + 6x + 9 \geq 0 =\ \textgreater \ (x + 3)^{2} \geq 0 \\

=\ \textgreater \ \ x\ apartine \ R \ (orice\ numar\ ridicat\ la\ patrat\ este\ \geq \ ca\ 0) [/tex]
x²+6x+10≥1 ⇔x²+6x+10-1≥0 ⇔x²+6x+9≥0 ⇔
(x+3)²≥0 adevarat oricare ar fi x∈R .