Đáp án:
${{B}_{2}}=3+2\sqrt{2}$
${{B}_{3}}=-3$
${{B}_{4}}=5+2\sqrt{6}$
Giải thích các bước giải:
${{B}_{2}}=\dfrac{\sqrt{2}+1}{\sqrt{2}-1}$
${{B}_{2}}=\dfrac{{{\left( \sqrt{2}+1 \right)}^{2}}}{\left( \sqrt{2}-1 \right)\left( \sqrt{2}+1 \right)}$
${{B}_{2}}=\dfrac{2+2\sqrt{2}+1}{2-1}$
${{B}_{2}}=3+2\sqrt{2}$
${{B}_{3}}=\left( 1-\sqrt{7} \right).\dfrac{\sqrt{7}+7}{2\sqrt{7}}$
${{B}_{3}}=\left( 1-\sqrt{7} \right).\dfrac{\sqrt{7}\left( 1+\sqrt{7} \right)}{2\sqrt{7}}$
${{B}_{3}}=\dfrac{\left( 1-\sqrt{7} \right)\left( 1+\sqrt{7} \right)}{2}$
${{B}_{3}}=\dfrac{1-7}{2}$
${{B}_{3}}=-3$
${{B}_{4}}=\dfrac{\sqrt{5+2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}$
${{B}_{4}}=\dfrac{\sqrt{{{\left( \sqrt{3}+\sqrt{2} \right)}^{2}}}}{\sqrt{3}-\sqrt{2}}$
${{B}_{4}}=\dfrac{\left| \sqrt{3}+\sqrt{2} \right|}{\sqrt{3}-\sqrt{2}}$
${{B}_{4}}=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
${{B}_{4}}=\dfrac{{{\left( \sqrt{3}+\sqrt{2} \right)}^{2}}}{\left( \sqrt{3}-\sqrt{2} \right)\left( \sqrt{3}+\sqrt{2} \right)}$
${{B}_{4}}=\dfrac{3+2\sqrt{6}+2}{3-2}$
${{B}_{4}}=5+2\sqrt{6}$
