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\(A=\dfrac{\sqrt{10+2\sqrt{21}}}{\sqrt{2}}+\dfrac{\sqrt{10-2\sqrt{21}}}{\sqrt{2}}-\dfrac{2}{\sqrt{2}}\sqrt{8-2\sqrt{7}}\)
\(A=\dfrac{\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}}{\sqrt{2}}+\dfrac{\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}{\sqrt{2}}-\dfrac{2}{\sqrt{2}}\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{7}+\sqrt{3}+\sqrt{7}-\sqrt{3}-2\sqrt{7}+2\right)=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
\(B=\dfrac{\sqrt[3]{2}\left(\sqrt[3]{2}+1+\sqrt[3]{2^2}\right)}{\sqrt[3]{4}+\sqrt[3]{2}+1}=\dfrac{\sqrt[3]{2}\left(\sqrt[3]{4}+\sqrt[3]{2}+1\right)}{\sqrt[3]{4}+\sqrt[3]{2}+1}=\sqrt[3]{2}\)
a ) \(\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)
b ) \(\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)
c ) \(\sqrt{21+4\sqrt{5}}=\sqrt{\left(2\sqrt{5}+1\right)^2}=2\sqrt{5}+1\)
d ) \(\sqrt{11+4\sqrt{7}}=\sqrt{\left(\sqrt{7}+2\right)^2}=\sqrt{7}+2\)
a) \(\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)
b) \(\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)
c) \(\sqrt{21+4\sqrt{5}}=\sqrt{\left(2\sqrt{5}+1\right)^2}=2\sqrt{5}+1\)
d) \(\sqrt{11+4\sqrt{7}}=\sqrt{\left(\sqrt{7}+2\right)^2}=\sqrt{7}+2\)
a
\(\sqrt{5-2\sqrt{6}}=\sqrt{\sqrt{2}^2-2.\sqrt{2}.\sqrt{3}+\sqrt{3}^2}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)
b,
\(\sqrt{3-2\sqrt{2}}=\sqrt{\sqrt{2}^2-2.\sqrt{2}.1+1^2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)
c,\(\sqrt{11+4\sqrt{7}}=\sqrt{11+2\sqrt{28}}=\sqrt{\sqrt{7}^2+2.\sqrt{7}.\sqrt{4}+\sqrt{4}^2}\)
\(=\sqrt{\left(\sqrt{7}+\sqrt{4}\right)^2}=\sqrt{7}+\sqrt{4}\)
a)\(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{10\left(4-\sqrt{15}\right)}+\sqrt{6\left(4-\sqrt{15}\right)}\)
\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(3-\sqrt{15}\right)^2}\)
\(=5-\sqrt{15}+\sqrt{15}-3\)
\(=2\)
b) \(2\left(\sqrt{10}-\sqrt{2}\right)\left(4+\sqrt{6-2\sqrt{5}}\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{\left(1-\sqrt{5}\right)^2}\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{5}-1\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=6\sqrt{10}+2\sqrt{50}-6\sqrt{2}-2\sqrt{10}\)
\(=6\sqrt{10}+10\sqrt{2}-6\sqrt{2}-2\sqrt{10}\)
\(=4\sqrt{10}+4\sqrt{2}\)
c) \(\left(\sqrt{7}+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)
\(=\left(\sqrt{7}+\sqrt{14}\right)\sqrt{\left(\sqrt{2}-\sqrt{7}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{14}\right)\left(\sqrt{7}-\sqrt{2}\right)\)
\(=7\sqrt{7}-7\sqrt{2}+\sqrt{98}-\sqrt{28}\)
\(=7\sqrt{7}-7\sqrt{2}+7\sqrt{2}-2\sqrt{7}\)
\(=5\sqrt{7}\)
d) \(\sqrt{\dfrac{289+4\sqrt{72}}{16}}\)
\(=\sqrt{\dfrac{289+42\sqrt{2}}{16}}\)
\(=\dfrac{\sqrt{289+42\sqrt{2}}}{\sqrt{4^2}}\)
\(=\dfrac{\sqrt{\left(1+12\sqrt{2}\right)^2}}{4}\)
\(=\dfrac{1+12\sqrt{2}}{4}\)
e) \(\left(\sqrt{21}+7\right)\sqrt{10-2\sqrt{21}}\)
\(=\left(\sqrt{21}+\sqrt{7}\right)\sqrt{\left(\sqrt{3}-\sqrt{7}\right)^2}\)
\(=\left(\sqrt{21}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{147}-\sqrt{63}+7-\sqrt{21}\)
\(=7\sqrt{3}-\sqrt{63}+7-\sqrt{21}\)
f) bạn xem đề lại nhé
a) \(\sqrt{7+3\sqrt{5}}=\sqrt{\left(\sqrt{\frac{5}{2}}+\sqrt{\frac{9}{2}}\right)^2}=\left|\sqrt{\frac{5}{2}}+\sqrt{\frac{9}{2}}\right|=\sqrt{\frac{5}{2}}+\sqrt{\frac{9}{2}}\)
b) \(\sqrt{238-30\sqrt{13}}=\sqrt{\left(\sqrt{225}-\sqrt{13}\right)^2}=\left|25-\sqrt{13}\right|=25-\sqrt{13}\)
c) \(\sqrt{118+28\sqrt{10}}=\sqrt{\left(\sqrt{20}+\sqrt{98}\right)^2}=\left|2\sqrt{5}+7\sqrt{2}\right|=2\sqrt{5}+7\sqrt{2}\)
(Nhớ k cho mình với nhá!)
\(a,\sqrt{3-\sqrt{5}}=\frac{\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\frac{\sqrt{5}-1}{\sqrt{2}}=\frac{\sqrt{10}-\sqrt{2}}{2}\)
\(b,\sqrt{4+\sqrt{7}}=\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{7}+1}{\sqrt{2}}=\frac{\sqrt{14}+\sqrt{2}}{2}\)
\(c,\sqrt{5+\sqrt{21}}=\frac{\sqrt{30+6\sqrt{21}}}{\sqrt{6}}=\frac{\sqrt{21}+3}{\sqrt{6}}\)