GIÚP MK NHANH NHA

\(\sqrt{16-6\sqrt{7}}=\sqrt{9-2.3\sqrt{7}+7}=\sqrt{\left(3-\sqrt{7}\right)^2}=3-\sqrt{7};\sqrt{10-2\sqrt{21}}=\sqrt{3-2\sqrt{3}\sqrt{7}+7}=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\Rightarrow\sqrt{16-6\sqrt{7}}+\sqrt{10-2\sqrt{21}}=3-\sqrt{3}\)

\(\sqrt{2-2.\frac{1}{2}\sqrt{2}+\frac{1}{4}}.\sqrt{8-2.2\sqrt{2}.\frac{1}{4}+\frac{1}{16}}=\sqrt{\left(\sqrt{2}-\frac{1}{2}\right)^2}\sqrt{\left(2\sqrt{2}-\frac{1}{4}\right)^2}\)

\(=\left(\sqrt{2}-\frac{1}{2}\right)\left(2\sqrt{2}-\frac{1}{4}\right)=\frac{33-10\sqrt{2}}{8}\)

\(\sqrt{2+2\sqrt{2}+1}.4\sqrt{\frac{288+2\sqrt{288}+1}{16}}=\sqrt{\left(\sqrt{2}+1\right)^2}.4\sqrt{\frac{\left(12\sqrt{2}+1\right)^2}{4^2}}\)

\(=\left(\sqrt{2}+1\right)\left(12\sqrt{2}+1\right)=25+13\sqrt{2}\)

\(\sqrt{28-10\sqrt{3}}=\sqrt{25-2.5\sqrt{3}+3}=\sqrt{\left(5-\sqrt{3}\right)^2}=5-\sqrt{3}\)

\(\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{\left(2\sqrt{2}-1\right)^2}}{2}=\frac{2\sqrt{2}-1}{2}\)

\(\sqrt{\frac{129+16\sqrt{2}}{16}}=\sqrt{\frac{\left(8\sqrt{2}+1\right)^2}{16}}=\frac{8\sqrt{2}+1}{4}\)

\(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)

\(\sqrt{\frac{289+4\sqrt{72}}{16}}=\frac{\sqrt{\left(12\sqrt{2}+1\right)^2}}{4}=\frac{12\sqrt{2}+1}{4}\)

\(\sqrt{8+2\sqrt{15}}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)

Học dỏi nha :)) 
~ Good luck ~

\(\sqrt{\frac{289+4\sqrt{72}}{16}}+\sqrt{\frac{129}{16}+\sqrt{2}}\)

\(=\sqrt{\frac{288+2\times12\sqrt{2}+1}{4^2}}+\sqrt{\frac{128+2\sqrt{12}+1}{4^2}}\)

\(=\sqrt{\frac{\left(\sqrt{288}+1\right)^2}{4^2}}+\sqrt{\frac{\left(\sqrt{128}+1\right)^2}{4^2}}\)

\(=\frac{\sqrt{288}+1}{4}+\frac{\sqrt{128}+1}{4}\)

\(=\frac{12\sqrt{2}+8\sqrt{2}+2}{4}\)

\(=\frac{1+10\sqrt{2}}{2}\)

a) Ta có: \(\frac{7\sqrt{2}+2\sqrt{7}}{\sqrt{14}}-\frac{5}{\sqrt{7}+\sqrt{5}}\)

\(=\frac{\sqrt{14}\left(\sqrt{7}+\sqrt{2}\right)}{\sqrt{14}}-\frac{5\left(\sqrt{7}-\sqrt{5}\right)}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)

\(=\frac{2\left(\sqrt{7}+\sqrt{2}\right)-5\left(\sqrt{7}-\sqrt{5}\right)}{2}\)

\(=\frac{2\sqrt{7}+2\sqrt{2}-5\sqrt{7}+5\sqrt{5}}{2}\)

\(=\frac{2\sqrt{2}-3\sqrt{7}+5\sqrt{5}}{2}\)

b) Ta có: \(\frac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)

\(=\frac{\sqrt{2}\left(6+2\sqrt{5}\right)}{4\sqrt{2}+\sqrt{2}\cdot\sqrt{6+2\sqrt{5}}}+\frac{\sqrt{2}\left(6-2\sqrt{5}\right)}{4\sqrt{2}-\sqrt{2}\cdot\sqrt{6-2\sqrt{5}}}\)

\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{2}\cdot\sqrt{\left(\sqrt{5}+1\right)^2}}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{2}\cdot\sqrt{\left(\sqrt{5}-1\right)^2}}\)

\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{2}\cdot\left|\sqrt{5}+1\right|}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{2}\cdot\left|\sqrt{5}-1\right|}\)

\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{2}\left(\sqrt{5}+1\right)}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{2}\cdot\left(\sqrt{5}-1\right)}\)(Vì \(\sqrt{5}>1>0\))

\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{10}+\sqrt{2}}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{10}+\sqrt{2}}\)

\(=\frac{6\sqrt{2}+2\sqrt{10}}{5\sqrt{2}+\sqrt{10}}+\frac{6\sqrt{2}-2\sqrt{10}}{5\sqrt{2}-\sqrt{10}}\)

\(=\frac{6+2\sqrt{5}}{5+\sqrt{5}}+\frac{6-2\sqrt{5}}{5-\sqrt{5}}\)

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

\(=\frac{\sqrt{5}+1+\sqrt{5}-1}{\sqrt{5}}\)

\(=\frac{2\sqrt{5}}{\sqrt{5}}=2\)

c) Đặt \(A=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\)

Ta có: \(A=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\)

\(\Leftrightarrow A^3=32-12\cdot\left(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\right)\)

\(=32-12A\)

\(\Leftrightarrow A^3+12A-32=0\)

\(\Leftrightarrow A^3-2A^2+2A^2-4A+16A-32=0\)

\(\Leftrightarrow A^2\left(A-2\right)+2A\left(A-2\right)+16\left(A-2\right)=0\)

\(\Leftrightarrow\left(A-2\right)\left(A^2+2A+16\right)=0\)

mà \(A^2+2A+16>0\)

nên A-2=0

hay A=2

Vậy: \(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}=2\)

a, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9}{4}-\frac{4\sqrt{2}}{4}}=\sqrt{\frac{9-4\sqrt{2}}{4}}\)

\(=\sqrt{\frac{\left(2\sqrt{2}\right)^2-4\sqrt{2}+1}{4}}=\sqrt{\frac{\left(2\sqrt{2}-1\right)^2}{4}}=\frac{2\sqrt{2}-1}{2}\)

b, \(\sqrt{\frac{129}{16}+\sqrt{2}}=\sqrt{\frac{129+16\sqrt{2}}{16}}=\sqrt{\frac{\left(8\sqrt{2}\right)^2+16\sqrt{2}+1}{16}}\)

\(=\sqrt{\frac{\left(8\sqrt{2}+1\right)^2}{16}}=\frac{8\sqrt{2}+1}{4}\)

\(b,\frac{2+\sqrt{3}}{1-\sqrt{4-2\sqrt{3}}}+\frac{2-\sqrt{3}}{1+\sqrt{4+2\sqrt{3}}}\)

\(=\frac{2+\sqrt{3}}{1-\sqrt{3-2\sqrt{3}+1}}+\frac{2-\sqrt{3}}{1+\sqrt{3+2\sqrt{3}+1}}\)

\(=\frac{2+\sqrt{3}}{1-\sqrt{\left(\sqrt{3}-1\right)^2}}+\frac{2-\sqrt{3}}{1+\sqrt{\left(\sqrt{3}+1\right)^2}}\)

\(=\frac{2+\sqrt{3}}{1-\left(\sqrt{3}-1\right)}+\frac{2-\sqrt{3}}{1+\sqrt{3}+1}\)

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

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

\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}\)

\(=14\)

\(a,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+4+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)

\(=\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\frac{\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2}{\sqrt{2}+\sqrt{3}+2}\)

\(=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)

\(=1+\sqrt{2}\)

Bài 2:

a)

\(\sqrt{9-\sqrt{17}}-\sqrt{9+\sqrt{17}}=\sqrt{\frac{18-2\sqrt{17}}{2}}-\sqrt{\frac{18+2\sqrt{17}}{2}}\)

\(=\sqrt{\frac{17+1-2\sqrt{17}}{2}}-\sqrt{\frac{17+1+2\sqrt{17}}{2}}=\sqrt{\frac{(\sqrt{17}-1)^2}{2}}-\sqrt{\frac{(\sqrt{17}+1)^2}{2}}\)

\(=\frac{\sqrt{17}-1}{\sqrt{2}}-\frac{\sqrt{17}+1}{\sqrt{2}}=-\sqrt{2}\)

b)

\(2\sqrt{2}(\sqrt{3}-2)+(1+2\sqrt{2})^2-2\sqrt{6}\)

\(=2\sqrt{6}-4\sqrt{2}+(1+4\sqrt{2}+8)-2\sqrt{6}\)

\(=1+8=9\)

Bài 1:

a)

\(\frac{\sqrt{6}+\sqrt{16}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{6}+4}{2(\sqrt{3}+\sqrt{7})}=\frac{1}{2}.\frac{(\sqrt{6}+4)(\sqrt{7}-\sqrt{3})}{(\sqrt{3}+\sqrt{7})(\sqrt{7}-\sqrt{3})}\)

\(=\frac{(4+\sqrt{6})(\sqrt{7}-\sqrt{3})}{8}\)

b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{16}-\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{(\sqrt{2}+1)(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)