Question

Va rog help la acest exercitiu dau 10 puncte si coroana

Answer 1

 

$\displaystyle\\a)\\\\\left(\frac{\sqrt{3}-\sqrt{2}}{2}\right)^2\cdot\left(\frac{\sqrt{3}+\sqrt{2}}{3}\right)^2+\frac{35}{36}=\\\\\\=\left[\left(\frac{\sqrt{3}-\sqrt{2}}{2}\right)\cdot\left(\frac{\sqrt{3}+\sqrt{2}}{3}\right)\right]^2+\frac{35}{36}=\\\\\\=\left[\left(\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)  }{2\cdot3}\right)\right]^2+\frac{35}{36}=$

$\displaystyle\\=\left(\frac{3-2}{2\cdot3}\right)^2+\frac{35}{36}=\\\\\\=\left(\frac{1}{6}\right)^2+\frac{35}{36}=\\\\\\=\frac{1}{3}+\frac{35}{36}=\frac{1+35}{36}=\frac{36}{36}=\boxed{\bf1}$

$\displaystyle\\b)\\\\\\\left(\frac{5}{3\sqrt{5}-2\sqrt{11}}\right)^3\cdot \left(\frac{9\sqrt{5}-6\sqrt{11}}{15}\right)^3 \cdot \left(-\frac{2\sqrt{3} }{3\sqrt{2} }   \right)^3+\frac{2\sqrt{6} }{9}=\\\\\\=\left(\frac{5\Big[9\sqrt{5}-6\sqrt{11}\Big]}{15\Big(3\sqrt{5}-2\sqrt{11}\Big)}\right)^3 \cdot \left(-\frac{2\sqrt{3} }{3\sqrt{2} }   \right)^3+\frac{2\sqrt{6} }{9}=$

$\displaystyle\\=\left(\frac{5\Big[3\Big(3\sqrt{5}-2\sqrt{11}\Big)\Big]}{15\Big(3\sqrt{5}-2\sqrt{11}\Big)}\right)^3 \cdot \left(-\frac{2\sqrt{3} }{3\sqrt{2} }   \right)^3+\frac{2\sqrt{6} }{9}=\\\\\\=\left(\frac{5\cdot3}{15}\right)^3 \cdot \left(-\frac{2\sqrt{3} }{3\sqrt{2} }   \right)^3+\frac{2\sqrt{6} }{9}=\\\\\\=\left(\frac{15}{15}\right)^3 \cdot \left(-\frac{2\sqrt{3} }{3\sqrt{2} }   \right)^3+\frac{2\sqrt{6} }{9}=\\\\\\=\left(-\frac{2\sqrt{3} }{3\sqrt{2} }   \right)^3+\frac{2\sqrt{6} }{9}=$

$\displaystyle\\=\left(-\frac{2\sqrt{3}\cdot \sqrt{2}}{3\sqrt{2} \cdot \sqrt{2}}   \right)^3+\frac{2\sqrt{6} }{9}=\\\\\\=\left(-\frac{2\sqrt{6}}{3 \cdot2}   \right)^3+\frac{2\sqrt{6} }{9}=\\\\\\=\left(-\frac{\sqrt{6}}{3}   \right)^3+\frac{2\sqrt{6} }{9}=\\\\\\=-\frac{6\sqrt{6}}{27}+\frac{3\cdot2\sqrt{6}}{3\cdot9}=\\\\\\=-\frac{6\sqrt{6}}{27}+\frac{6\sqrt{6}}{27}=\boxed{\bf0}$