Đáp án:
Câu 79: $ A=\dfrac{\sqrt{6}(3+\sqrt{2}+\sqrt3)(8-6\sqrt2)}{-8}$
Câu 80:
Giải thích các bước giải:
Câu 79:
Ta có:
$A=\dfrac{\sqrt{6}}{3+\sqrt{2}-\sqrt3}$
$\to A=\dfrac{\sqrt{6}(3+\sqrt{2}+\sqrt3)}{(3+\sqrt{2}-\sqrt3)(3+\sqrt{2}+\sqrt3)}$
$\to A=\dfrac{\sqrt{6}(3+\sqrt{2}+\sqrt3)}{(3+\sqrt{2})^2-(\sqrt3)^2}$
$\to A=\dfrac{\sqrt{6}(3+\sqrt{2}+\sqrt3)}{8+6\sqrt{2}}$
$\to A=\dfrac{\sqrt{6}(3+\sqrt{2}+\sqrt3)(8-6\sqrt2)}{(8+6\sqrt{2})(8-6\sqrt2)}$
$\to A=\dfrac{\sqrt{6}(3+\sqrt{2}+\sqrt3)(8-6\sqrt2)}{-8}$
Câu 80:
Ta có:
$B=\dfrac{2\sqrt3}{\sqrt5+\sqrt6+\sqrt7}$
$\to B=\dfrac{2\sqrt3(\sqrt5+\sqrt6-\sqrt7)}{(\sqrt5+\sqrt6+\sqrt7)(\sqrt5+\sqrt6-\sqrt7)}$
$\to B=\dfrac{2\sqrt3(\sqrt5+\sqrt6-\sqrt7)}{(\sqrt5+\sqrt6)^2-(\sqrt7)^2}$
$\to B=\dfrac{2\sqrt3(\sqrt5+\sqrt6-\sqrt7)}{4+2\sqrt{30}}$
$\to B=\dfrac{\sqrt3(\sqrt5+\sqrt6-\sqrt7)}{2+\sqrt{30}}$
$\to B=\dfrac{\sqrt3(\sqrt5+\sqrt6-\sqrt7)(2-\sqrt{30})}{(2+\sqrt{30})(2-\sqrt{30})}$
$\to B=\dfrac{\sqrt3(\sqrt5+\sqrt6-\sqrt7)(2-\sqrt{30})}{2^2-(\sqrt{30})^2}$
$\to B=\dfrac{\sqrt3(\sqrt5+\sqrt6-\sqrt7)(2-\sqrt{30})}{-26}$
