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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 )
\(a,=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\\ b,=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\\ c,=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
1) ĐKXĐ: \(x\ge5\)
2) ĐKXĐ: \(\left[{}\begin{matrix}x< -2\\x>2\end{matrix}\right.\)
5) ĐKXĐ: \(\left[{}\begin{matrix}x\le2\\x\ge3\end{matrix}\right.\)
\(VT=\left[\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right]\cdot\left(\sqrt{7}-\sqrt{5}\right)\\ =\left(-\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\\ =-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2=VP\)
\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}=\left(-\sqrt{7}-\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2\)
\(A=\dfrac{8+2\sqrt{15}+\sqrt{21}+\sqrt{35}}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{3}+\sqrt{5}+\sqrt{7}\right)+\sqrt{3}\left(\sqrt{3}+\sqrt{5}+\sqrt{7}\right)}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{3}+\sqrt{5}+\sqrt{7}\right)}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\sqrt{5}+\sqrt{3}\)
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