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a/ Ta có: \(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(3+\sqrt{5}\right)^2}\)
\(=3-\sqrt{5}+3+\sqrt{5}=6\)
b/ \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\sqrt{5}-2-\sqrt{5}-2=-4\)
\(\sqrt{\left(6-2\sqrt{5}\right)^3}-\sqrt{\left(6-2\sqrt{5}\right)^3}=0\)
\(\sqrt{14+6\sqrt{5}}-\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}}=5\)
\(\sqrt{14+6\sqrt{5}}-\sqrt{\frac{\sqrt{5-2}}{\sqrt{5}+2}}=5\)
\(\sqrt{14+6\sqrt{5}}-\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}}=\sqrt{9+3.2\sqrt{5}+5}-\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}}=\sqrt{3^2+3.2\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}}=\sqrt{\left(3+\sqrt{5}\right)^2}-\sqrt{\frac{\left(\sqrt{5}-2\right)^2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}=3+\sqrt{5}-\sqrt{\frac{\left(\sqrt{5}-2\right)^2}{5-2^2}}=3+\sqrt{5}-\sqrt{\left(\sqrt{5}-2\right)^2}=3+\sqrt{5}-\left(\sqrt{5}-2\right)\left(2=\sqrt{4}< \sqrt{5}\right)=3+\sqrt{5}-\sqrt{5}+2=5\)
\(\sqrt{\left(\sqrt{5}+3\right)^2}+\sqrt{14-6\sqrt{5}}\)\(=\left|\sqrt{5}+3\right|+\sqrt{9-2.3\sqrt{5}+5}\)
\(=\sqrt{5}+3+\sqrt{\left(3-\sqrt{5}\right)^2}\) \(=\sqrt{5}+3+\left|3-\sqrt{5}\right|\)
\(=\sqrt{5}+3+3-\sqrt{5}=6\) ( do \(3-\sqrt{5}>0\))
a.\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}=\left|\sqrt{3}+2\right|=\sqrt{3}+2\)
b.\(\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-2\right|=\sqrt{5}-2\)
c.\(\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}+3\right)^2}=\left|\sqrt{5}+3\right|=\sqrt{5}+3\)
d.\(\sqrt{17-12\sqrt{2}}=\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)
Ta có :
a)\(\left(2\sqrt{5}-\sqrt{7}\right)\left(2\sqrt{5}-\sqrt{7}\right)=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)
b)\(\left(5\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{3}-5\sqrt{2}\right)=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
c)\(\sqrt{9+4\sqrt{5}}=\sqrt{2^2+2.2.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2+\sqrt{5}\right|=2+\sqrt{5}\)
Công thức viết khó đọc quá. Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn.
a) \(\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)
\(=\sqrt{14}\cdot\sqrt{5-\sqrt{21}}+\sqrt{6}\cdot\sqrt{5-\sqrt{21}}\)
\(=\sqrt{14\cdot\left(5-\sqrt{21}\right)}+\sqrt{6\cdot\left(5-\sqrt{21}\right)}\)
\(=\sqrt{70-14\sqrt{21}}+\sqrt{30-6\sqrt{21}}\)
\(=\sqrt{7^2-2\cdot7\cdot\sqrt{21}+\left(\sqrt{21}\right)^2}+\sqrt{\left(\sqrt{21}\right)^2-2\cdot3\cdot\sqrt{21}+3^2}\)
\(=\sqrt{\left(7-\sqrt{21}\right)^2}+\sqrt{\left(\sqrt{21}-3\right)^2}\)
\(=\left|7-\sqrt{21}\right|+\left|\sqrt{21}-3\right|\)
\(=7-\sqrt{21}+\sqrt{21}-3\)
\(=4\)
b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left[4\cdot\left(\sqrt{10}-\sqrt{6}\right)+\sqrt{15}\cdot\left(\sqrt{10}-\sqrt{6}\right)\right]\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)
\(=\sqrt{10\cdot\left(4-\sqrt{15}\right)}+\sqrt{6\cdot\left(4-\sqrt{15}\right)}\)
\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{5^2-2\cdot5\cdot\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2\cdot3\cdot\sqrt{15}+3^2}\)
\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)
\(=\left|5-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)
\(=5-\sqrt{15}+\sqrt{15}-3\)
\(=2\)
\(P=\sqrt{14+6\sqrt{5}}+\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{3^2+6\sqrt{5}+\sqrt{5}^2}+\sqrt{3^2-6\sqrt{5}+\sqrt{5}^2}\)
\(=\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(=\left|3+\sqrt{5}\right|+\left|3-\sqrt{5}\right|\)
\(=3+\sqrt{5}+3-\sqrt{5}\)
\(=6\)