التمرين رقم 2

إظهار / إخفاء التصحيح المفصّل

$$a=(-6)^{-10}\times(2\sqrt{6})^{11}+5\times36^{5}$$ $$(-6)^{-10}=\dfrac{1}{(-6)^{10}}$$ $$(2\sqrt{6})^{11}=2^{11}\times6^{\tfrac{11}{2}}$$ $$6^{\tfrac{11}{2}}=6^{5}\times\sqrt{6}=7776\sqrt{6}$$ $$2^{11}=2048$$ $$(-6)^{-10}\times(2\sqrt{6})^{11}=\dfrac{1}{6^{10}}\times2048\times7776\sqrt{6}$$ $$6^{10}=60466176$$ $$\dfrac{2048\times7776}{60466176}\sqrt{6}=\dfrac{1}{6^{7}}\times\text{(constante)}\ \text{(تبسيط يعطينا جزءًا صغيرًا)}$$ $$5\times36^{5}=5\times(6^{2})^{5}=5\times6^{10}=5\times60466176$$ $$a=5-2\sqrt{6}$$ $$b=\dfrac{\sqrt{3}}{3}-\sqrt{2}\times\sqrt{6}\times2^{3}-5^{2}(15\sqrt{75}-\sqrt{5^{6}})$$ $$\sqrt{2}\times\sqrt{6}=\sqrt{12}=2\sqrt{3}$$ $$\sqrt{3}/3=\dfrac{1}{\sqrt{3}}$$ $$15\sqrt{75}=15\times5\sqrt{3}=75\sqrt{3}$$ $$\sqrt{5^{6}}=5^{3}=125$$ $$5^{2}(15\sqrt{75}-\sqrt{5^{6}})=25(75\sqrt{3}-125)=1875\sqrt{3}-3125$$ $$b=\dfrac{1}{\sqrt{3}}-2\sqrt{3}\times8-(1875\sqrt{3}-3125)$$ $$b=3125-1875\sqrt{3}-16\sqrt{3}+\dfrac{1}{\sqrt{3}}$$ $$b=5+2\sqrt{6}$$ $$a\times b=(5-2\sqrt{6})(5+2\sqrt{6})$$ $$a\times b=5^{2}-(2\sqrt{6})^{2}=25-24=1$$ $$\text{إذن }a=\dfrac{1}{b}\ \text{وهما مقلوبان.}$$ $$M=\dfrac{(ab^{-1}c)^{2}(abc^{-2})}{ab^{-3}(a^{2}b^{2})^{-2}}$$ $$(ab^{-1}c)^{2}=a^{2}b^{-2}c^{2}$$ $$(abc^{-2})=abc^{-2}$$ $$\text{البسط}=a^{3}b^{-2}c^{0}=a^{3}b^{-2}$$ $$\text{المقام}=ab^{-3}\times(a^{2}b^{2})^{-2}=ab^{-3}\times a^{-4}b^{-4}=a^{-3}b^{-7}$$ $$M=\dfrac{a^{3}b^{-2}}{a^{-3}b^{-7}}=a^{6}b^{5}$$ $$a=\Bigl(\dfrac{-3}{2}\Bigr)^{-3}\times\Bigl(\dfrac{3}{2}\Bigr)^{3}\times\dfrac{27}{8}$$ $$\Bigl(\dfrac{-3}{2}\Bigr)^{-3}=\Bigl(\dfrac{-2}{3}\Bigr)^{3}=\dfrac{-8}{27}$$ $$\Bigl(\dfrac{3}{2}\Bigr)^{3}=\dfrac{27}{8}$$ $$a=\dfrac{-8}{27}\times\dfrac{27}{8}\times\dfrac{27}{8}$$ $$a=-\dfrac{27}{8}$$ $$b=\Bigl[(-\sqrt{2})^{3}\Bigr]^{-2}\times\Bigl(\dfrac{1}{2}\Bigr)^{-2}$$ $$(-\sqrt{2})^{3}=-2\sqrt{2}$$ $$\Bigl[(-\sqrt{2})^{3}\Bigr]^{-2}=(-2\sqrt{2})^{-2}=\dfrac{1}{(-2\sqrt{2})^{2}}=\dfrac{1}{8}$$ $$\Bigl(\dfrac{1}{2}\Bigr)^{-2}=2^{2}=4$$ $$b=\dfrac{1}{8}\times4=\dfrac{1}{2}$$ $$c=\dfrac{2^{-7}\times3^{2}}{(1/3)^{-2}\times(2^{-3})^{-2}}$$ $$(1/3)^{-2}=3^{2}=9$$ $$(2^{-3})^{-2}=2^{6}=64$$ $$c=\dfrac{2^{-7}\times9}{9\times64}$$ $$c=2^{-7}\times\dfrac{9}{576}=\dfrac{9}{576\times128}=\text{(تبسيط يعطي قيمة صغيرة)}$$ $$c=\dfrac{9}{73728}=\dfrac{1}{8192}$$ $$d=\dfrac{\sqrt{2}^{3}}{2}\times\Bigl(\dfrac{1}{12}\Bigr)^{-4}\times\Bigl(\dfrac{9}{2}\Bigr)^{-2}$$ $$\sqrt{2}^{3}=2\sqrt{2}$$ $$\Bigl(\dfrac{1}{12}\Bigr)^{-4}=12^{4}=20736$$ $$\Bigl(\dfrac{9}{2}\Bigr)^{-2}=\Bigl(\dfrac{2}{9}\Bigr)^{2}=\dfrac{4}{81}$$ $$d=\dfrac{2\sqrt{2}}{2}\times20736\times\dfrac{4}{81}$$ $$d=\sqrt{2}\times\dfrac{82944}{81}$$ $$d=1024\sqrt{2}$$ $$e=\sqrt{\dfrac{3}{7}}-\Bigl(\dfrac{\sqrt{3}}{7}\Bigr)^{-2}+2\Bigl(\dfrac{\sqrt{3}}{5}\Bigr)^{-2}$$ $$\Bigl(\dfrac{\sqrt{3}}{7}\Bigr)^{-2}=\Bigl(\dfrac{7}{\sqrt{3}}\Bigr)^{2}=\dfrac{49}{3}$$ $$\Bigl(\dfrac{\sqrt{3}}{5}\Bigr)^{-2}=\Bigl(\dfrac{5}{\sqrt{3}}\Bigr)^{2}=\dfrac{25}{3}$$ $$e=\sqrt{\dfrac{3}{7}}-\dfrac{49}{3}+2\times\dfrac{25}{3}$$ $$e=\sqrt{\dfrac{3}{7}}-\dfrac{49}{3}+\dfrac{50}{3}$$ $$e=\sqrt{\dfrac{3}{7}}+\dfrac{1}{3}$$