Xem lại câu c) và d) 

b: =căn 10-3+4-căn 10=1

a: \(=\sqrt{11-4\sqrt{6}+\sqrt{15}}\)

Bình phương lên nhé tự hỏi tự trl

\(=\left|3-\sqrt{10}\right|+\sqrt{\left(\sqrt{10}-1\right)^2}\\ =\sqrt{10}-3+\sqrt{10}-1=\sqrt{10}-4\)

\(\sqrt{28-10\sqrt{3}}\\ =\sqrt{3-10\sqrt{3}+25}\\ =\sqrt{\left(\sqrt{3}-5\right)^2}\\ =\left|\sqrt{3}-5\right|\\ =5-\sqrt{3}\)

\(\sqrt{41+12\sqrt{5}}\\ =\sqrt{5+12\sqrt{5}+36}\\ =\sqrt{\left(\sqrt{5}+6\right)}\\ =\left|\sqrt{5}+6\right|\\ =\sqrt{5}+6\)

\(\sqrt{32-10\sqrt{7}}\\ =\sqrt{7-10\sqrt{7}+25}\\ =\sqrt{\left(\sqrt{7}-5\right)^2}\\ =\left|\sqrt{7}-5\right|\\ =5-\sqrt{7}\)

\(\sqrt{11-4\sqrt{7}}\\ =\sqrt{7-4\sqrt{7}+4}\\ =\sqrt{\left(\sqrt{7}-2\right)^2}=\left|\sqrt{7}-2\right|\\ =\sqrt{7}-2\)

câu a coi lai có sai sót j ko

\(A=\sqrt{11-2\sqrt{10}}+\sqrt{9-2\sqrt{4}}-\sqrt{10}-\sqrt{7}\)

\(=\sqrt{\left(\sqrt{10}-1\right)^2}+\sqrt{5}-\sqrt{10}-\sqrt{7}=\sqrt{10}-1+\sqrt{5}-\sqrt{10}-\sqrt{7}\)

\(=\sqrt{5}-\sqrt{7}-1\)

\(\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)

\(=\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{\left(\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}\right)\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)}{\left(\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}\right)\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)}\)

\(=\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{5+3\sqrt{2}-\left(5-3\sqrt{2}\right)}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{3+\sqrt{2}-\left(3-\sqrt{2}\right)}\)

\(=\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{6\sqrt{2}}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{2\sqrt{2}}\)

\(=\frac{\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{2\sqrt{2}}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{2\sqrt{2}}\) 

\(\frac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)

\(=\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)

\(=\frac{\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{2\sqrt{2}}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{2\sqrt{2}}\)

\(=\frac{10+2\sqrt{7}-6-2\sqrt{7}}{2\sqrt{2}}=\sqrt{2}\)  

\(\sqrt{\sqrt{11}+1}.\sqrt{\sqrt{11}-1}+\sqrt{10}=\sqrt{10}+\sqrt{10}=2\sqrt{10}\)

a) \(\sqrt{11+2\sqrt[]{18}}\)

\(=\sqrt{11+6\sqrt[]{2}}\)

\(=\sqrt{9+2.3\sqrt[]{2}+2}\)

\(=\sqrt{\left(3+\sqrt[]{2}\right)^2}=\left|3+\sqrt[]{2}\right|=3+\sqrt[]{2}\)

b) \(\sqrt[]{7+2\sqrt[]{10}}\)

\(=\sqrt[]{7+2\sqrt[]{5}.\sqrt[]{2}}\)

\(=\sqrt[]{5+2\sqrt[]{5}.\sqrt[]{2}+2}\)

\(=\sqrt[]{\left(\sqrt[]{5}+\sqrt[]{2}\right)^2}=\left|\sqrt[]{5}+\sqrt[]{2}\right|=\sqrt[]{5}+\sqrt[]{2}\)

c) \(\sqrt[]{7+4\sqrt[]{3}}\)

\(=\sqrt[]{4+2.2\sqrt[]{3}+3}\)

\(=\sqrt[]{\left(2+\sqrt[]{3}\right)^2}=\left|2+\sqrt[]{3}\right|=2+\sqrt[]{3}\)

d) \(\sqrt[]{16-2\sqrt[]{55}}\) \(\left(12\rightarrow16\right)\)

\(=\sqrt[]{11-2\sqrt[]{5}.\sqrt[]{11}+5}\)

\(=\sqrt[]{\left(\sqrt[]{11}-\sqrt[]{5}\right)^2}==\left|\sqrt[]{11}-\sqrt[]{5}\right|=\sqrt[]{11}-\sqrt[]{5}\left(\sqrt[]{11}>\sqrt[]{5}\right)\)