câu b,c ạ, giúp em với ,...

Đáp án:

b) $\sqrt{(1-\sqrt{2})^2}+\sqrt{11-6\sqrt{2}}=2$

c) $\left(\dfrac{10}{\sqrt{6}-1}-\dfrac{\sqrt{12}+\sqrt{30}}{\sqrt{5}+\sqrt{2}}\right).\sqrt{5-2\sqrt{6}}=5\sqrt{2}-4\sqrt{3}$

Giải thích các bước giải:

b)

$\sqrt{(1-\sqrt{2})^2}+\sqrt{11-6\sqrt{2}}\\=|1-\sqrt{2}|+\sqrt{(3-\sqrt{2})^2}\\=\sqrt{2}-1+|3-\sqrt{2}|\\=\sqrt{2}-1+3-\sqrt{2}\\=2$

c)

$\left(\dfrac{10}{\sqrt{6}-1}-\dfrac{\sqrt{12}+\sqrt{30}}{\sqrt{5}+\sqrt{2}}\right).\sqrt{5-2\sqrt{6}}\\=\left(\dfrac{10(\sqrt{6}+1)}{(\sqrt{6}-1)(\sqrt{6}+1)}-\dfrac{\sqrt{6}(\sqrt{5}+2)}{\sqrt{5}+2}\right)\sqrt{(\sqrt{3}-\sqrt{2})^2}\\=(2(\sqrt{6}+1)-\sqrt{6}).(\sqrt{3}-\sqrt{2})\\=(\sqrt{6}-2)(\sqrt{3}-2)\\=\sqrt{2}(\sqrt{3}-\sqrt{2})^2=\sqrt{2}(5-2\sqrt{6})\\=5\sqrt{2}-4\sqrt{3}$