Ta có: \(\dfrac{8+2\sqrt{15}+\sqrt{21}+\sqrt{35}}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)

\(=\dfrac{\left(\sqrt{3}+\sqrt{5}\right)^2+\sqrt{7}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)

\(=1+\sqrt{3}+\sqrt{5}\)

\(B=\left(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+11\right)\)

\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)

\(=6-121=-115\) là số nguyên (đpcm)

b) Ta có: \(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)

=6-121=-115

\(Q=\sqrt{\sqrt{5}-1}\left(\sqrt{8-\sqrt{5}+2\sqrt{5\sqrt{5}-3}}-\sqrt{7-\sqrt{20}}\right)\)

\(\Rightarrow\)\(Q^2=\left(\sqrt{5}-1\right)\left(8-\sqrt{5}+2\sqrt{5\sqrt{5}-3}+7-\sqrt{20}-2\sqrt{\left(7-\sqrt{20}\right)\left(8-\sqrt{5}+2\sqrt{5\sqrt{5}-3}\right)}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{\left(7-2\sqrt{5}\right)\left(8-\sqrt{5}\right)+2\left(7-2\sqrt{5}\right)\sqrt{5\sqrt{5}-3}}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{66-23\sqrt{5}+2\left(7-2\sqrt{5}\right)\sqrt{5\sqrt{5}-3}}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{\left(49-28\sqrt{5}+20\right)+2\left(7-2\sqrt{5}\right)\sqrt{5\sqrt{5}-3}+\left(5\sqrt{5}-3\right)}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{\left(7-2\sqrt{5}\right)^2+2\left(7-2\sqrt{5}\right)\sqrt{5\sqrt{5}-3}+\left(5\sqrt{5}-3\right)}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{\left(7-2\sqrt{5}+\sqrt{5\sqrt{5}-3}\right)^2}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\left(7-2\sqrt{5}+\sqrt{5\sqrt{5}-3}\right)\right)\)

\(=\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)\)\(=4\)

\(\Rightarrow Q^2=4\) \(\Rightarrow Q\) nguyên 

Trả lời:

\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-12\sqrt{5}+9}}}\)

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

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

\(A=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

\(A=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\)

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

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

\(A=\sqrt{1}\)

\(A=1\)

\(B=\frac{\left(5+2\sqrt{6}\right).\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{\left(3+2\sqrt{6}+2\right).\left(49-20\sqrt{6}\right).\sqrt{3-2\sqrt{6}+2}}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2.\left(49-20\sqrt{6}\right).\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2.\left(49-20\sqrt{6}\right).\left(\sqrt{3}-\sqrt{2}\right)}{9\sqrt{33}-11\sqrt{2}}\)

\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right).\left(\sqrt{3}-\sqrt{2}\right).\left(\sqrt{3}+\sqrt{2}\right).\left(49-20\sqrt{6}\right)}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{\left(3-2\right).\left(49\sqrt{3}-60\sqrt{2}+49\sqrt{2}-40\sqrt{3}\right)}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{1.\left(9\sqrt{3}-11\sqrt{2}\right)}{9\sqrt{3}-11\sqrt{2}}\)

\(B=1\)

a) Ta có: \(\sqrt{29-12\sqrt{5}}=\sqrt{20-12\sqrt{5}+9}=\sqrt{\left(2\sqrt{5}-3\right)^2}\)

\(=\left|2\sqrt{5}-3\right|=2\sqrt{5}-3\)

\(\Rightarrow\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{3-\left(2\sqrt{5}-3\right)}=\sqrt{3-2\sqrt{5}+3}\)

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

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

\(\Leftrightarrow A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)

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

\(\left\{{}\begin{matrix}a=\dfrac{35}{49}=\dfrac{5}{7}\\b=\sqrt{\dfrac{5^2}{7^2}}=\dfrac{5}{7}\\c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\dfrac{5+35}{7+49}=\dfrac{5}{7}\\d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\dfrac{5-35}{7-49}=\dfrac{5}{7}\end{matrix}\right.\)

\(\Rightarrow a=b=c=d=\dfrac{5}{7}\)

\(a=\dfrac{35}{49};b=\dfrac{5}{7}\\ c,=\dfrac{5+35}{7+49}=\dfrac{12}{14}=\dfrac{6}{7}\\ d,=\dfrac{5-35}{7-49}\)

Áp dụng t/c dtsbn:

\(\dfrac{5}{7}=\dfrac{35}{49}=\dfrac{5+35}{7+49}=\dfrac{5-35}{7-49}\) hay \(a=b=c=d\)

em tham khảo

\(a,=\dfrac{\sqrt{x}-8+5}{\sqrt{x}-8}=1+\dfrac{5}{\sqrt{x}-8}\in Z\\ \Leftrightarrow\sqrt{x}-8\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{3;7;9;13\right\}\\ \Leftrightarrow x\in\left\{9;49;81;169\right\}\left(tm\right)\\ b,=\dfrac{\sqrt{x}-2+7}{\sqrt{x}-2}=1+\dfrac{7}{\sqrt{x}-2}\in Z\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(7\right)=\left\{-1;1;7\right\}\left(\sqrt{x}-2>-2\right)\\ \Leftrightarrow\sqrt{x}\in\left\{1;3;9\right\}\\ \Leftrightarrow x\in\left\{1;9;81\right\}\\ c,=\dfrac{2\left(\sqrt{x}+3\right)+2}{\sqrt{x}+3}=2+\dfrac{2}{\sqrt{x}+3}\in Z\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(2\right)=\varnothing\left(\sqrt{x}+3>3\right)\\ \Leftrightarrow x\in\varnothing\)

\(a,=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\\ b,=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\\ c,=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)