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צפה בגרסה המלאה : משוואות מעריכיות



דניאלייי
07-09-2010, 15:59
אשמח לעזרה.
תודה אאש ושנה טובה!

Imageshack - scan0002og.jpg (http://img843.imageshack.us/i/scan0002og.jpg/)

avishay12456
07-09-2010, 16:58
lתרגיל 119:

http://latex.codecogs.com/gif.latex?25^{x-1}-6\cdot%205^{x-1.5}=-1%20\\25^x\cdot%2025^{-1}-6\cdot%205^{-2}\cdot%205^{0.5}\cdot%205^x=-1%20\\(5^2)^x\cdot%20\frac{1}{25}-6\cdot%20\frac{1}{25}\cdot%20\sqrt{5}\cdot%205^x=-1%20\\(5^x)^2\cdot%20\frac{1}{25}-\frac{6\cdot%20\sqrt{5}}{25}\cdot%205^x+1=0/\cdot%2025%20\\(5^x)^2-6\cdot%20\sqrt{5}\cdot%205^x+25=0\%20\%20\%20\%205 ^x=t%20\\t^2-6\sqrt{5}\%20t+25=0%20\\\\t_{1,2}=\frac{6\sqrt{5}\ pm\sqrt{(-6\sqrt{5})^2-4\cdot%201\cdot%2025}}{2\cdot%201}=\frac{6\sqrt{5} \pm\sqrt{80}}{2}=\frac{\sqrt{180}\pm\sqrt{80}}{2}= %20\\\frac{\sqrt{4\cdot%2045}\pm\sqrt{4\cdot%2020} }{2}=\frac{\sqrt{4}(\sqrt{45}\pm\sqrt{20})}{2}=\fr ac{2(\sqrt{5\cdot%209}\pm\sqrt{5\pm4})}{2}=%20\\\s qrt{5\cdot%209}\pm\sqrt{5\cdot%204}=(\sqrt{9}\pm\s qrt{4})\sqrt{5}=(3\pm2)\sqrt{5}%20\\t_1=(3+2)\sqrt {5}=5\sqrt{5}%20\\t_2=(3-2)\sqrt{5}=\sqrt{5}%20\\\\5^x=5\sqrt{5}%20\\\left% 20[%20(\sqrt{5})^2%20\right%20]^x=(\sqrt{5})^2\cdot%20\sqrt{5}%20\\(\sqrt{5})^{2x }=(\sqrt{5})^{2+1}%20\\2x=2+1%20\\2x=3/:2%20\\x=1.5%20\\\\5^x=\sqrt{5}%20\\\left%20[%20(\sqrt{5})^2%20\right%20]^x=\sqrt{5}%20\\(\sqrt{5})^{2x}=\sqrt{5}%20\\2x=1/:2%20\\x=0.5



תרגיל 121:
http://latex.codecogs.com/gif.latex?36^x-6\cdot%2018^x+8\cdot%209^x=0%20\\(6^2)^x-6\cdot%206^x\cdot%203^x+8\cdot%20(3^2)^x=0%20\\6^{ 2x}-6^{x+1}\cdot%203^x+8\cdot%203^{2x}=0%20\\2^{2x}\cd ot%203^{2x}-2^{x+1}\cdot%203^{x+1}\cdot%203^x+8\cdot%203^{2x}= 0%20\\2^{2x}\cdot%203^{2x}-2^{x+1}\cdot%203^{2x+1}+8\cdot%203^{2x}=0/:3^{2x}%20\\2^{2x}-2^{x+1}\cdot%203+8=0%20\\(2^x)^2-2\cdot%202^x\cdot%203+8=0\%20\%20\%20\%202^x=t%20\ \t^2-6t+8=0%20\\t_{1,2}=\frac{6\pm\sqrt{(-6)^2-4\cdot%201\cdot%208}}{2\cdot%201}=\frac{6\pm\sqrt{ 4}}{2}=\frac{6\pm2}{2}%20\\t_1=\frac{6+2}{2}=\frac {8}{2}=4%20\\t_2=\frac{6-2}{2}=\frac{4}{2}=2%20\\\\\\2^x=4%20\\2^x=2^2%20\\ x=2%20\\\\\\2^x=2%20\\x=1