[tex](\sqrt{3+2\sqrt{6}+2})^{x}+(\sqrt{3-2\sqrt{6}+2})^{x}=10[/tex] [tex](\sqrt{(\sqrt{3}+\sqrt{2})^2})^{x}+(\sqrt{(\sqrt{3}-\sqrt{2})^2})^{x}=10[/tex] [tex](\sqrt{3}+\sqrt{2})^{x}+(\sqrt{3}-\sqrt{2})^{x}=10[/tex] Toliau nzn kaip :/
Mirtise (+3503)
na ir aš panašiai dariau... tik reikia dėt modulį dar [tex]\left ( \left | \sqrt{2}+\sqrt{3} \right | \right )^{x}+\left ( \left | \sqrt{2}-\sqrt{3} \right | \right )^{x}= 10[/tex] tada naikini modulį [tex]\left ( \sqrt{2}+\sqrt{3} \right )^{x}+\left (- \sqrt{2}+\sqrt{3} \right )^{x}= 10[/tex] ;D
aidas (+63)
MilkhaterĮsiveskite keitinį √3 + √2 = t tada √3 - √2 = 1/t
Gal detaliau, kaip is √3 + √2 = t i si isvedei? √3 - √2 = 1/t