Question

Exercițiul 19 punctele A si B

Answer 1

$\displaystyle 19~a)~(2\sqrt{7}-3\sqrt{3})^2-(3\sqrt{3}-2\sqrt{5})^2=\\ \\=(2\sqrt{7} )^2-2\cdot 2\sqrt{7}\cdot 3\sqrt{3}+(3\sqrt{3} )^2-((3\sqrt{3} )^2-2 \cdot 3\sqrt{3}\cdot 2\sqrt{5}+(2\sqrt{5} )^2)=\\ \\=28-12\sqrt{21}+27-(27-12\sqrt{15} +20)=\\ \\=28-12\sqrt{21}+27-27+12\sqrt{15} -20=\mathbf{8+12\sqrt{15} -12\sqrt{21}}$

$\displaystyle b)~a\sqrt{5}-b\sqrt{7} =\sqrt{55-12\sqrt{21}}+\sqrt{47-12\sqrt{15}}\\ \\ \\\sqrt{55-12\sqrt{21}}=\sqrt{\frac{55+\sqrt{55^2-12^2\cdot21}}{2}}  -\sqrt{\frac{55-\sqrt{55^2-12^2\cdot21}}{2} }=\\ \\ =\sqrt{\frac{55+\sqrt{3025-144\cdot21}}{2}} -\sqrt{\frac{55-\sqrt{3025-144\cdot21}}{2}}=\\ \\=\sqrt{\frac{55+\sqrt{3025-3024}}{2}}-\sqrt{\frac{55-\sqrt{3025-3024}}{2}}=$

$\displaystyle =\sqrt{\frac{55+\sqrt{1}}{2}}- \sqrt{\frac{55-\sqrt{1}}{2}}=\sqrt{\frac{55+1}{2}}-\sqrt{\frac{55-1}{2}}=\sqrt{\frac{56}{2}}-\sqrt{\frac{54}{2}}=\\ \\=\sqrt{28}-\sqrt{27} =2\sqrt{7} -3\sqrt{3}$

$\displaystyle\sqrt{47-12\sqrt{15}}=\sqrt{\frac{47+\sqrt{47^2-12^2\cdot15}}{2}}-\sqrt{\frac{47-\sqrt{47^2-12^2\cdot15}}{2}} =\\ \\ =\sqrt{\frac{47+\sqrt{2209-144\cdot15}}{2} }-\sqrt{\frac{47-\sqrt{2209-144\cdot15} }{2}}=\\ \\ =\sqrt{\frac{47+\sqrt{2209-2160}}{2} }-\sqrt{\frac{47-\sqrt{2209-2160}}{2}}=\\ \\ =  \sqrt{\frac{47+\sqrt{49}}{2}}-\sqrt{\frac{47-\sqrt{49}}{2}} =\sqrt{\frac{47+7}{2}}-\sqrt{\frac{47-7}{2}}=\sqrt{\frac{54}{2}}-\sqrt{\frac{40}{2}}=\\ \\ =\sqrt{27}-\sqrt{20}=3\sqrt{3} -2\sqrt{5}$

$a\sqrt{5} -b\sqrt{7}= \sqrt{55-12\sqrt{21}}+\sqrt{47-12\sqrt{15}} \\ \\ a\sqrt{5} -b\sqrt{7} =2\sqrt{7} -3\sqrt{3} +3\sqrt{3} -2\sqrt{5} \\ \\ a\sqrt{5} -b\sqrt{7} =-2\sqrt{5} +2\sqrt{7} \\ \\ \mathbf{a=-2;~b=-2}$