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ajutatima va rog cu ex 3 si 4 multumesc

Ajutatima Va Rog Cu Ex 3 Si 4 Multumesc class=

Răspuns :

3) [tex]x = \frac{4(\sqrt{6}-\sqrt{2})}{6-2} + \sqrt{2}(\sqrt{2}-\sqrt{3})+\frac{2}{\sqrt{2}} =\ \textgreater \ x = \frac{4(\sqrt{6}-\sqrt{2})}{4}+2-\sqrt{6}+\frac{2\sqrt{2}}{2} \\ =\ \textgreater \ x = \sqrt{6}-\sqrt{2}+2-\sqrt{6}+\sqrt{2} =\ \textgreater \ x = 2[/tex]
[tex]y=|\sqrt{5}-3|+2|1-\sqrt{5}|+|-1| =\ \textgreater \ y = 3-\sqrt{5}+2(\sqrt{5}-1)+1 =\ \textgreater \ \\ =\ \textgreater \ y=3-\sqrt{5}+2\sqrt{5}-2+1 =\ \textgreater \ y = 2 + \sqrt{5}[/tex]
4) [tex]-1 \leq \frac{x+2}{2} \leq 5 | * 2 =\ \textgreater \ -2 \leq x+2 \leq 10 |-2 =\ \textgreater \ -4 \leq x \leq 8 =\ \textgreater \ \\ =\ \textgreater \ A = [-4,8] \\ |x-2| \leq 6 =\ \textgreater \ -6 \leq x -2 \leq 6 | +2 =\ \textgreater \ -4 \leq x \leq 8 =\ \textgreater \ B = [-4,8] \\ =\ \textgreater \ A = B[/tex]

Doar ex 3 l-am facut pas cu pas
Vezi imaginea IOAN97