\(\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}=\frac{\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right).\sqrt{\sqrt{5}-1}}{\sqrt{\sqrt{5}+1}.\sqrt{\sqrt{5}-1}}=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{\sqrt{5-1}}\)\(=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}=\frac{\sqrt{2}.\left(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}\right)}{2\sqrt{2}}=\frac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}}{2\sqrt{2}}\)

\(=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}}{2\sqrt{2}}=\frac{\left|\sqrt{5}+1\right|+\left|3-\sqrt{5}\right|}{2\sqrt{2}}=\frac{\sqrt{5}+1+3-\sqrt{5}}{2\sqrt{2}}\)

\(=\frac{4}{2\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)

Lại có: \(\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|=\sqrt{2}-1\)

\(\Rightarrow\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}=\sqrt{2}-\left(\sqrt{2}-1\right)=1\)

2. 

a,  Với m\(=1\Rightarrow x^2-x=0\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

b. Ta có \(\Delta=b^2-4ac=\left(-m\right)^2-4\left(m-1\right)=m^2-4m+4=\left(m-2\right)^2\ge0\)

\(\Rightarrow\)phương trình luôn có 2 nghiệm \(x_1,x_2\)

c, Theo hệ thức Viet ta có \(\hept{\begin{cases}x_1+x_2=m\\x_1.x_2=m-1\end{cases}}\)

A=\(\frac{2.x_1x_2+3}{x_1^2+x_2^2+2\left(1+x_1x_2\right)}=\frac{2.x_1x_2+3}{\left(x_1+x_2\right)^2-2x_1x_2+2+2x_1x_2}\)

\(=\frac{2x_1x_2+3}{\left(x_1+x_2\right)^2+2}=\frac{2m+1}{m^2+2}=\frac{\left(m^2+2\right)-\left(m^2-2m+1\right)}{m^2+2}\)

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

Ta thấy \(\frac{-\left(m-1\right)^2}{m^2+2}\le0\Rightarrow1+\frac{-\left(m-1\right)^2}{m^2+2}\le1\)

\(\Rightarrow MaxA=1\)

Dấu bằng xảy ra\(\Leftrightarrow\) \(m-1=0\Leftrightarrow m=1\)

Từ A kẻ đường thẳng vuông góc với AN cắt CD tại E

Ta có AB=mAD nên \(\frac{AB}{AD}=m\)

Xét \(\Delta ABM\)và  \(\Delta ADE\)có :

góc ABM = góc ADE =90

góc BAM =góc FAD (cùng phụ với góc DAN )

\(\Rightarrow\Delta ABM~\Delta ADF\left(g.g\right)\)\(\Rightarrow\frac{AM}{AF}=\frac{AB}{AD}=m\)\(\Rightarrow\frac{1}{AF}=\frac{m}{AM};\frac{1}{AD}=\frac{m}{AB}\)

Tam giác AFN VUÔNG TẠI A CÓ \(AD⊥FN\)\(\Rightarrow\frac{1}{AD^2}=\frac{1}{AF^2}+\frac{1}{AN^2}\)

                                                      HAY     \(\left(\frac{m}{AB}\right)^2=\left(\frac{m}{AM}\right)^2+\frac{1}{AN^2}\Rightarrow\frac{m^2}{AB^2}=\frac{m^2}{AM^2}+\frac{1}{AN^2}\left(đpcm\right)\)

https://olm.vn/hoi-dap/question/667592.html