Question

Daca a = 1/√3 - 2/-√12 + 1/√27 si b =

√6/√8+√54/√32 atunci a × b este :

Answer 1

 

$\displaystyle\bf\\a=\frac{1}{\sqrt3}-\frac{2}{-\sqrt{12}}+\frac{1}{\sqrt{27}}\\\\a=\frac{1}{\sqrt3}+\frac{2}{\sqrt{12}}+\frac{1}{\sqrt{27}}\\\\a=\frac{1}{\sqrt3}+\frac{2}{\sqrt{4\times3}}+\frac{1}{\sqrt{9\times3}}\\\\a=\frac{1}{\sqrt3}+\frac{2}{2\sqrt{3}}+\frac{1}{3\sqrt{3}}\\\\\\a=\frac{1}{\sqrt3}+\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{3}}\\\\\\a=\frac{\sqrt3}{3}+\frac{\sqrt3}{3}+\frac{\sqrt3}{3\times3}\\\\\\a=\frac{3\sqrt3}{9}+\frac{3\sqrt3}{9}+\frac{\sqrt3}{9}\\\\\\\boxed{\bf~a=\frac{7\sqrt3}{9}}$

.

$\displaystyle\bf\\b=\frac{\sqrt{6}}{\sqrt{8}}+\frac{\sqrt{54}}{\sqrt{32}}\\\\\\b=\frac{\sqrt{2\times3}}{\sqrt{4\times2}}+\frac{\sqrt{27\times2}}{\sqrt{16\times2}}\\\\\\b=\frac{\sqrt{2}\times\sqrt{3}}{\sqrt{4}\times\sqrt{2}}+\frac{\sqrt{27}\times\sqrt{2}}{\sqrt{16}\times\sqrt{2}}\\\\\\b=\frac{\sqrt{3}}{\sqrt{4}}+\frac{\sqrt{27}}{\sqrt{16}}\\\\\\b=\frac{\sqrt{3}}{\sqrt{4}}+\frac{\sqrt{9\times3}}{\sqrt{16}}\\\\\\b=\frac{\sqrt{3}}{\sqrt{4}}+\frac{\sqrt{9}\times\sqrt{3}}{\sqrt{16}}$

.

$\displaystyle\bf\\b=\frac{\sqrt{3}}{2}+\frac{3\sqrt{3}}{4}\\\\\\b=\frac{2\sqrt{3}}{4}+\frac{3\sqrt{3}}{4}\\\\\\\boxed{\bf~b=\frac{5\sqrt{3}}{4}}\\\\\\\\a\times b=\frac{7\sqrt{3}}{9}\times\frac{5\sqrt{3}}{4}\\\\\\a\times b=\frac{7\times5\times\sqrt{3}\times\sqrt{3}}{9\times4}\\\\\\a\times b=\frac{35\times3}{36}\\\\\\\boxed{\bf~a\times b=\frac{35}{12}}\\\\\\\boxed{\bf~a\times b=2\frac{11}{12}}$