b) Ta có: \(\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}+\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)

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

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

\(=\dfrac{3+3}{2}=\dfrac{6}{2}=3\)

=\(\frac{1}{\sqrt{7-2\sqrt{6}_{ }}+1}+\frac{1}{\sqrt{7+2\sqrt{6}}+1}\)

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

=\(\frac{1}{\sqrt{6}}+\frac{1}{\sqrt{6}+2}\)

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

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

\(\sqrt{7+\sqrt{24}=\sqrt{7+2\sqrt{6}}=\sqrt{\left(\sqrt{6}+1\right)^2}}\)

mình cũng làm vậy nhưng ko ra kết quả 

\(\frac{1}{\sqrt{7-\sqrt{24}}+1}+\frac{1}{\sqrt{7+\sqrt{24}}+1}\)

\(=\frac{1}{\sqrt{7-2\sqrt{6}}+1}+\frac{1}{\sqrt{7+2\sqrt{6}}+1}\)

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

\(=\frac{1}{\sqrt{6}-1+1}+\frac{1}{\sqrt{6}+1+1}\)

\(=\frac{1}{\sqrt{6}}+\frac{1}{\sqrt{6}+2}=\frac{\sqrt{6}+2+\sqrt{6}}{6+2\sqrt{6}}=\frac{2\sqrt{6}+2}{6+2\sqrt{6}}\)

ta có :

\(P=\frac{\sqrt{x}+4}{1-7\sqrt{x}}+\frac{\sqrt{x}-2}{\sqrt{x}+1}+\frac{24\sqrt{x}}{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)}\)

\(\frac{-\left(\sqrt{x}+4\right)\left(\sqrt{x}+1\right)+\left(\sqrt{x}-2\right)\left(7\sqrt{x}-1\right)+24\sqrt{x}}{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)}=\frac{6x+4\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)}\)

\(=\frac{6\sqrt{x}+2}{7\sqrt{x}-1}\)

Để \(P\ge-6\Leftrightarrow\frac{6\sqrt{x}+2}{7\sqrt{x}-1}\ge-6\Leftrightarrow\frac{48\sqrt{x}-4}{7\sqrt{x}-1}\ge0\)

\(\Leftrightarrow\orbr{\begin{cases}0\le\sqrt{x}\le\frac{1}{12}\\\sqrt{x}>\frac{1}{7}\end{cases}}\Leftrightarrow\orbr{\begin{cases}0\le x\le\frac{1}{144}\\x>\frac{1}{49}\end{cases}}\)

\(A=\frac{\sqrt{1}+\sqrt{2}}{1-2}-\frac{\sqrt{2}+\sqrt{3}}{2-3}+\frac{\sqrt{3}+\sqrt{4}}{3-4}-...-\frac{\sqrt{24}+\sqrt{25}}{24-25}\)

\(=-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{24}+\sqrt{25}\)

\(=-\sqrt{1}+\sqrt{25}\)

\(=-1+5\)

\(=4.\)

Nhân liên hiệp ta được :

\(\frac{\sqrt{1}+\sqrt{2}}{\left(\sqrt{1}-\sqrt{2}\right)\left(\sqrt{1}+\sqrt{2}\right)}+\frac{\sqrt{2}+\sqrt{3}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{\sqrt{24}+\sqrt{25}}{\left(\sqrt{24}-\sqrt{25}\right)\left(\sqrt{24}+\sqrt{25}\right)}\)

\(=\frac{\sqrt{1}+\sqrt{2}}{1-2}+\frac{\sqrt{2}+\sqrt{3}}{2-3}+...+\frac{\sqrt{24}+\sqrt{25}}{24-25}\)

\(=-\sqrt{1}-\sqrt{2}-\sqrt{2}-\sqrt{3}-....-\sqrt{24}-\sqrt{25}\)

\(=-\left[\frac{\left(\sqrt{25}+\sqrt{1}\right).25}{2}+\frac{\left(\sqrt{24}+\sqrt{2}\right).23}{2}\right]\)

\(=...\)