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\(B=\dfrac{21}{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}\right)^2-3\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}\right)^2-15\sqrt{15}\)
\(=\dfrac{21}{2}\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-3\left(\sqrt{3}-1+\sqrt{5}+1\right)^2-15\sqrt{15}\)
\(=\dfrac{21}{2}\left(\sqrt{3}+\sqrt{5}\right)^2-3\left(\sqrt{3}+\sqrt{5}\right)^2-15\sqrt{15}\)
\(=\dfrac{15}{2}\left(8+2\sqrt{15}\right)-15\sqrt{15}\)
\(=60+15\sqrt{15}-15\sqrt{15}=60\)
\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{14-6\sqrt{3}}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{\frac{20+4\sqrt{3}-10\sqrt{3}-6}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{\frac{4\left(5+\sqrt{3}\right)-2\sqrt{3}\left(5+\sqrt{3}\right)}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{\frac{\left(4-2\sqrt{3}\right)\left(5+\sqrt{3}\right)}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}=\left(\sqrt{3}+1\right)\sqrt{3-2\sqrt{3}+1}=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=3-1=2\Rightarrow A=2\)
Bài 1:
Xét tử số:
\(\sqrt{14+6\sqrt{5}}-\sqrt{14-6\sqrt{5}}=\sqrt{3^2+5+2.3\sqrt{5}}-\sqrt{3^2+5-2.3\sqrt{5}}\)
\(=\sqrt{(3+\sqrt{5})^2}-\sqrt{(3-\sqrt{5})^2}=3+\sqrt{5}-(3-\sqrt{5})=2\sqrt{5}\)
Xét mẫu số:
\(\sqrt{(\sqrt{5}+1)\sqrt{6-2\sqrt{5}}}=\sqrt{(\sqrt{5}+1)\sqrt{5+1-2\sqrt{5}}}=\sqrt{(\sqrt{5}+1)\sqrt{(\sqrt{5}-1)^2}}\)
\(=\sqrt{(\sqrt{5}+1)(\sqrt{5}-1)}=\sqrt{4}=2\)
Do đó: $A=\frac{2\sqrt{5}}{2}=\sqrt{5}$
\(=\left(\sqrt{3}+1\right)\sqrt{\frac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\frac{70-14\sqrt{3}-30\sqrt{3}+18}{25-\sqrt{3}^2}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\frac{88-44\sqrt{3}}{22}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
mình không viết lại đề nha
\(=\sqrt{\frac{\left(\sqrt{3}+1\right)^2.\left(14-6\sqrt{3}\right)}{5+\sqrt{3}}}\)
\(=\sqrt{\frac{\left(3+2\sqrt{3}+1\right).\left(14-6\sqrt{3}\right)}{5+\sqrt{3}}}\)
\(=\sqrt{\frac{\left(4+2\sqrt{3}\right).\left(14-6\sqrt{3}\right)}{5+\sqrt{3}}}\)
\(=\sqrt{\frac{56-24\sqrt{3}+28\sqrt{3}-36}{5+\sqrt{3}}}\)
\(=\sqrt{\frac{20+4\sqrt{3}}{5+\sqrt{3}}}\)
\(=\sqrt{\frac{\left(20+4\sqrt{3}\right).\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right).\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{\frac{100-20\sqrt{3}+20\sqrt{3}-12}{5^2-\sqrt{3}^2}}\)
\(=\sqrt{\frac{88}{25-3}}\)
\(=\sqrt{\frac{88}{22}}\)
\(=\sqrt{4}\)
\(=2\)
HỌC TỐT !!!
Bài 2:
Ta có: \(B=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)
\(=\frac{\sqrt{\sqrt{5}-1}\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)}{2}-\sqrt{2-2\cdot\sqrt{2}\cdot1+1}\)
\(=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\frac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}}{2\sqrt{2}}-\left(\sqrt{2}-1\right)\)
\(=\frac{\sqrt{5}+1+3-\sqrt{5}}{2\sqrt{2}}-\sqrt{2}+1\)
\(=\frac{4}{2\sqrt{2}}-\sqrt{2}+1\)
\(=\sqrt{2}-\sqrt{2}+1\)
=1
câu 3: C = \(\frac{\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)}{\left(\text{4+\sqrt{15}}\right)\left(\sqrt{10-\sqrt{6}}\right)\sqrt{4-\sqrt{15}}}\)
\(=\frac{\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}.\sqrt{3+\sqrt{5}}}{\sqrt{4+\sqrt{15}}.\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}}\)
=\(\frac{\sqrt{9-\left(\sqrt{5}\right)^2}\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}}{\sqrt{16-\left(\sqrt{15}\right)^2}.\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4+\sqrt{15}}}\)
\(=\frac{2\left(\sqrt{30+10\sqrt{5}}-\sqrt{6+2\sqrt{5}}\right)}{\sqrt{40+10\sqrt{15}}-\sqrt{24-6\sqrt{15}}}\)
\(=2.\frac{\left(\sqrt{5}+5\right)-\left(\sqrt{5}+1\right)}{\left(\sqrt{15}+5\right)-\left(\sqrt{15}+3\right)}\)
= 4
\(A=\frac{\left(\sqrt{3}+1\right)\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}=\frac{4\left(11\sqrt{3}-11\right)}{25^2-\left(\sqrt{3}\right)^2}=\frac{44\left(\sqrt{3}-1\right)}{22}=2\sqrt{3}-2\)
\(\frac{4\left(11\sqrt{3}-11\right)}{25^2-\left(\sqrt{3}\right)^2}\) Thay Bằng:
\(\frac{4\left(11\sqrt{3}-11\right)}{5^2-\left(\sqrt{3}\right)^2}\)
Cảm ơn bạn!