Ta có :

\(b^2=\left(3+\sqrt{6+\sqrt{7+\sqrt{2}}}\right)\left(3-\sqrt{6+\sqrt{7+\sqrt{2}}}\right)\)

\(b^2=9-\left(6+\sqrt{7+\sqrt{2}}\right)\)

\(b^2=3-\sqrt{7+\sqrt{2}}\)

\(\Rightarrow b=\sqrt{3-\sqrt{7+\sqrt{2}}}\)

Tích ab :

\(ab=\sqrt{2+\sqrt{2}}.\sqrt{3+\sqrt{7+\sqrt{2}}}.\sqrt{3-\sqrt{7+\sqrt{2}}}\)

\(ab=\sqrt{2+\sqrt{2}}.\left(9-7-\sqrt{2}\right)\)

\(ab=\sqrt{2+\sqrt{2}}.\left(2-\sqrt{2}\right)\)

P/s : làm được thế này thui . Sai bỏ qua

a) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

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

\(A=\frac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2}{\sqrt{x}}\)

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

b) \(A>\frac{1}{2}\)

\(\Leftrightarrow\frac{2}{\sqrt{x}+1}>\frac{1}{2}\)

\(\Leftrightarrow\frac{2}{\sqrt{x}+1}-\frac{1}{2}>0\)

\(\Leftrightarrow\frac{4-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}>0\)

\(\Leftrightarrow\frac{3-\sqrt{x}}{2\sqrt{x}+2}>0\)

\(\Leftrightarrow3-\sqrt{x}>0\)( \(2\sqrt{x}+2>0\)với mọi x lớn hơn hoặc bằng 0; x khác 4 )

\(\Leftrightarrow-\sqrt{x}>-3\)

\(\Leftrightarrow\sqrt{x}< 3\)

\(\Leftrightarrow x< 9\)

Vậy với x>9 ; \(x\ge0\); x khác 4 thì A>1/2

c) Ta có : \(B=\frac{7}{3}A\)

\(\Leftrightarrow B=\frac{14}{3\left(\sqrt{x}+2\right)}\)

\(\Leftrightarrow B=\frac{14}{3\sqrt{x}+6}\)

B là số nguyên

\(\Leftrightarrow3\sqrt{x}+6\inƯ\left(14\right)\)

Vì \(3\sqrt{x}+6>0\)với mọi \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

=> chỉ chọn giá trị dương

+) Bạn tự xét các trường hợp

Kết quả ra : \(\orbr{\begin{cases}x=\frac{1}{9}\\x=\frac{64}{9}\end{cases}}\)

Vậy ............

\(a,\frac{6}{4+\sqrt{4-2\sqrt{3}}}=\frac{6}{4+\sqrt{\sqrt{3}^2-2\sqrt{3}+\sqrt{1}^2}}\)

\(=\frac{6}{4+\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}}=\frac{6}{4+|\sqrt{3}-1|}=\frac{6}{3+\sqrt{3}}\)

\(=\frac{6}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{36}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{3}.\sqrt{12}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{12}}{\sqrt{3}+1}\)

\(d,\frac{1}{\sqrt{7-2\sqrt{10}}}+\frac{1}{\sqrt{7+2\sqrt{10}}}\)

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

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

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

\(=\frac{2\sqrt{5}}{\sqrt{5}^2-\sqrt{2}^2}=\frac{\sqrt{5.4}}{5-2}=\frac{\sqrt{20}}{3}\)

\(\sqrt{x+4\sqrt{x-1}+3}-\sqrt{4x+4\sqrt{x-1}-3}=1\)(đk:\(1\le x< 2\)) Lý do có điều kiện này là nhờ vào việc VT=1>0

\(\Leftrightarrow\sqrt{\left(x-1\right)+4\sqrt{x-1}+4}-\sqrt{4\left(x-1\right)+4\sqrt{x-1}+1}=1\)

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

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

\(\Leftrightarrow\sqrt{x-1}=0\)

\(\Leftrightarrow x=1\)(thõa mãn điều kiện)

Ta có : \(\sqrt{x+4\sqrt{x-1}+3}-\sqrt{4x+4\sqrt{x-1}-3}=1\) ( ĐK : \(x\ge1\) )

\(\Leftrightarrow\sqrt{\left(x-1\right)+4\sqrt{x-1}+4}-\sqrt{4.\left(x-1\right)+4.\sqrt{x-1}+1}=1\)

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

\(\Leftrightarrow\left|\sqrt{x-1}+2\right|-\left|2\sqrt{x-1}+1\right|=1\)

\(\Leftrightarrow\sqrt{x-1}+2-2\sqrt{x-1}-1=1\)

\(\Leftrightarrow\sqrt{x-1}=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\) ( Thỏa mãn )

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

              \(=8-2\sqrt{2+\sqrt{3}}-2\sqrt{3\left(2+\sqrt{2+\sqrt{3}}\right)\left(2-\sqrt{2+\sqrt{3}}\right)}\)

              \(=8-\frac{2}{\sqrt{2}}\sqrt{4+2\sqrt{3}}-2\sqrt{3\left(2^2-\sqrt{2+\sqrt{3}}^2\right)}\)       

              \(=8-\sqrt{2}\sqrt{\sqrt{3}^2+2\cdot1\sqrt{3}+1^2}-2\sqrt{3\left(4-2-\sqrt{3}\right)}\)

              \(=8-\sqrt{2}\sqrt{\left(\sqrt{3}+1\right)^2}-2\sqrt{3}\sqrt{2-\sqrt{3}}\)

               \(=8-\sqrt{2}\left(\sqrt{3}+1\right)-\frac{2\sqrt{3}}{\sqrt{2}}\sqrt{4-2\sqrt{3}}\)

               \(=8-\left(\sqrt{6}+\sqrt{2}\right)-\sqrt{6}\sqrt{\left(\sqrt{3}-1\right)^2}\)

               \(=8-\sqrt{6}-\sqrt{2}-\sqrt{6}\left(\sqrt{3}-1\right)\)

               \(=8-\sqrt{6}-\sqrt{2}-\sqrt{18}+\sqrt{6}\)

               \(=8-\sqrt{2}-\sqrt{18}\)

               \(=8-\sqrt{2}\left(3+1\right)=8-4\sqrt{2}\)

\(\Rightarrow x^4-16x^2=\left(8-4\sqrt{2}\right)^2-16\left(8-4\sqrt{2}\right)\)

                          \(=8^2+4^2\cdot\sqrt{2}^2-2\cdot8\cdot4\sqrt{2}-16\cdot8+16\cdot4\sqrt{2}\)

                          \(=64+32-64\sqrt{2}-128+64\sqrt{2}\)

                          \(=-32\)

         Vậy \(x^4-16x^2=-32\)

Tại hạ làm bừa có gì mong đạo hữu lượng thứ =))