Ta có: \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)

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

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

                                                                                    \(=2\sqrt{2}\)

\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

\(=\sqrt{9+2.3\sqrt{2}+2}-\sqrt{9-2.3\sqrt{2}+2}\)

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

\(=2\sqrt{2}\)

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

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

\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+\sqrt{3}}}\)\(+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}\)

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

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

Sai chỗ nào ý nhỉ

\(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{2^2+2.2.\sqrt{5}+\sqrt{5}^2}-\sqrt{2^2-2.2.\sqrt{5}+\sqrt{5}^2}\)

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

\(=|2+\sqrt{5}|-|2-\sqrt{5}|\)

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

\(=4\)

\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)

= \(\sqrt{7}-1-\left(1+\sqrt{7}\right)\)

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

= -2

1.  \(2\sqrt{5}-5\sqrt{20}+\sqrt{80}\)

= \(2\sqrt{5}-5.2\sqrt{5}+4\sqrt{5}\)

= \(2\sqrt{5}-10\sqrt{5}+4\sqrt{5}\)

= \(-4\sqrt{5}\)

2. B = \(\frac{1}{\sqrt{5}-2}-\sqrt{6-2\sqrt{5}}\)

       = \(\frac{1}{\sqrt{5}-2}-\sqrt{1^2-2\sqrt{5}+\sqrt{15}^2}\)

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

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

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

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

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

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

Thay vào A ta được:

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

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

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

Ta có \(\frac{a^3}{b^2+1}=\frac{a^3}{b^2+ab}=\frac{a^3}{b\left(a+b\right)}\)

Áp dụng BĐT cosi

\(\frac{a^3}{b\left(a+b\right)}+\frac{b}{2}+\frac{a+b}{4}\ge\frac{3}{2}a\)

TT \(\frac{b^3}{a\left(a+b\right)}+\frac{a}{2}+\frac{a+b}{4}\ge\frac{3}{2}b\)

=> \(VT\ge\frac{1}{2}\left(a+b\right)\ge\sqrt{ab}=1\)

Dấu bằng xảy ra khi a=b=1

Sửa đề \(A=\frac{x^2}{x+y^2}+\frac{y^2}{y+z^2}+\frac{z^2}{z+x^2}\)

Ta có \(\frac{x^2}{x+y^2}=\frac{x^2+xy^2-xy^2}{x+y^2}=x-\frac{xy^2}{x+y^2}\)

Mà \(x+y^2\ge2y\sqrt{x}\)

=> \(\frac{x^2}{x+y^2}\ge x-\frac{\sqrt{x}}{2}\ge\frac{3x}{4}-\frac{1}{4}\)

=> \(VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}=\frac{9}{4}-\frac{3}{4}=\frac{3}{2}\)

Min A=3/2 khi x=y=z=1