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1. Determinati numerele reale a,b daca (1-i[tex] \sqrt{3} [/tex])^2=a(1+i[tex] \sqrt{3} [/tex])+b
2. Fie [tex]a_n [/tex] n[tex] \geq [/tex]1 o progresie aritmetica. Daca [tex] a_{2}+a_{5}+a_{7}+a_{10}=4, calculati a_{1}+a_{2}+.... +a_{11}[/tex]
3. Rezolvati ecuatia 2[tex] sin^{2} [/tex]x-3cosx=0, x∈R
4. Aratati ca a=log in baza 1/[tex] \sqrt{2} [/tex]din2+log in baza[tex] \sqrt{2} [/tex]din4 este nr natural.

Răspuns :

Matepentrutotiid
1)[tex](1-i\sqrt{3})^2=a(1+i\sqrt{3})+b\\ 1-2\sqrt{3}i-3=a+a\sqrt{3}i+b\\ \left \{ {{-2\sqrt{3}=a\sqrt3}=>a=-2} \atop {1-3=a+b=>a+b=-2=>-2+b=-2=>b=0}} \right. [/tex]
2)[tex]a_2+a_5+a_7+a_{10}=a_1+r+a_1+4r+a_1+6r+a_1+9r=\\ 4a+1+20r=4|:4\\ a_1+5r=1\\ a_1+...+a_{11}=\frac{(a_1+a_{11})\cdot 11}{2}=\frac{(a_1+a_{1}+10r)\cdot 11}{2}=\\ =\frac{2(a_1+5r)\cdot 11}{2}=1 \cdot 11=11[/tex]
3)[tex]2(1-cos^2x)-3cosx=0\\ cos^2+3cosx-2=0\\ Notam \ \ cosx=t\\ 2t^2+3t-2=0\\ t_1=\frac{1}{2},t_2=-2\\ cosx=\frac{1}{2}=>x=\pm arcos\frac{1}{2}+2k\pi=>x=\pm\frac{\pi}{6}+2k\pi,k\in Z[/tex]
4)[tex] a=log_{ \frac{1}{ \sqrt{2} } 2}+log_{\sqrt{2}}4=\\ = \frac{log_22}{log_22^{-1/2}} +\frac{log_22^2}{log_22^{1/2}}=\\ = \frac{1}{-1/2} + \frac{2}{1/2}=-2+4=2 \in N [/tex]

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