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\(A=\sqrt{5+\sqrt{21}}+\sqrt{5-\sqrt{21}}\)
\(A^2=\left(\sqrt{5+\sqrt{21}}^2+2\sqrt{\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)}+\sqrt{5-\sqrt{21}}^2\right)\)
\(A^2=5+\sqrt{21}+\sqrt{4\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)}+5-\sqrt{21}\)
\(A^2=10+\sqrt{4.\left(25-21\right)}\)
\(A^2=10+\sqrt{4.4}=10+4=14\)
\(A=\sqrt{14}\)
Tương tự
a. \(=\sqrt{2}.\left(\sqrt{7}+\sqrt{8}\right)\sqrt{5-\sqrt{3}\sqrt{7}}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{3-2\sqrt{3}.\sqrt{7}+7}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
Rồi nhân ra. bạn làm tiếp nhé. Tuy nhiên minh nghĩ bạn bị nhầm đề. là \(\sqrt{6}\) chứ không phải căn 16
b. \(=\frac{5\left(\sqrt{21}+1\right)}{21-16}+\frac{\sqrt{3}.\sqrt{7}\left(\sqrt{3}-\sqrt{7}\right)}{-\left(\sqrt{3}-\sqrt{7}\right)}\)
\(=\sqrt{21}+4-\sqrt{21}=4\)
\(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)
\(\Leftrightarrow\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}=4\)
\(\Leftrightarrow\sqrt{25}-\sqrt{1}=4\Leftrightarrow5-1=4\)(đúng)
Vậy \(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)(đpcm)
\(M=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{11-6\sqrt{2}}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{2-6\sqrt{2}+9}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{\left(3-\sqrt{2}\right)^2}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+3-\sqrt{2}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{6}}\)
\(=\sqrt{16+32\sqrt{6}}\)
a) Ta có: \(A=\frac{8+2\sqrt{15}+\sqrt{21}+\sqrt{35}}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\frac{\left(\sqrt{3}+\sqrt{5}\right)^2+\sqrt{7}\cdot\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\frac{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}+\sqrt{7}\right)}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\sqrt{3}+\sqrt{5}\)
b) Ta có: \(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}\)
\(=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{4}+\sqrt{3}\right)\left(\sqrt{4}-\sqrt{3}\right)}+\frac{\sqrt{5}-\sqrt{4}}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+\frac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+2-\sqrt{3}+\sqrt{5}-2+\sqrt{6}-\sqrt{5}\)
\(=-1+\sqrt{6}\)
\(A=\sqrt{\frac{5+\sqrt{21}}{5-\sqrt{21}}}+\sqrt{\frac{5-\sqrt{21}}{5+\sqrt{21}}}\)
\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}\)
\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{4}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{4}}\)
\(=\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}=5\)
\(B=\sqrt{7+\sqrt{33}}+\sqrt{7-\sqrt{33}}\)
\(\Rightarrow\)\(\sqrt{2}B=\sqrt{14+2\sqrt{33}}+\sqrt{14-2\sqrt{33}}\)
\(=\sqrt{\left(\sqrt{11}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}\)
\(=\sqrt{11}+\sqrt{3}+\sqrt{11}-\sqrt{3}=2\sqrt{11}\)
\(\Rightarrow\)\(B=\sqrt{22}\)
cho mk hỏi căn viết thế nào