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a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}=\sqrt{16}-6+\sqrt{20}-\sqrt{5}=4-6+2\sqrt{5}-\sqrt{5}=\sqrt{5}-2\)
b) \(0,2\sqrt{\left(-10\right)^3.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=0,2\left|-10\right|\sqrt{3}+2\left|\sqrt{3}-\sqrt{5}\right|=0,2.10.\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)
c) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}=\left(\dfrac{1}{2}\sqrt{\dfrac{2}{4}}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{2}{3}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{27}{4}\sqrt{2}.8=54\sqrt{2}\)
d) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2.\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}=2\left(3-\sqrt{2}\right)+3\sqrt{2}-5=6-2\sqrt{2}+3\sqrt{2}-5=1+\sqrt{2}\)
\(\sqrt{\frac{3\sqrt{5}+1}{2\sqrt{5}-3}}\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}\)
?? :v
\(\sqrt{\frac{3\sqrt{5}+1}{2\sqrt{5}-3}}\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{\frac{\left(3\sqrt{5}+1\right)\left(2\sqrt{5}+3\right)}{\left(2\sqrt{5}-3\right)\left(2\sqrt{5}+3\right)}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{3+\sqrt{5}}.\sqrt{2}.\left(\sqrt{5}-1\right)\)
\(=\sqrt{6+2\sqrt{5}}.\left(\sqrt{5}-1\right)\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=4\)
Làm luôn nhé
\(2B=21.2\left[\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)\right]^2-2.15\sqrt{15}\)
\(2B=21\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-6\left(\sqrt{3}-1+\sqrt{5}-1\right)^2-30\sqrt{15}\)
\(2B=21\left(\sqrt{3}+\sqrt{5}\right)^2-6\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(8+2\sqrt{15}\right)-30\sqrt{15}\)
\(2B=120+30\sqrt{15}-30\sqrt{5}\)
\(2B=120\)
\(B=60\)
\(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
\(=\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{5}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
= 1
\(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\times\sqrt{3+\sqrt{5}}\times\sqrt{2}\left(\sqrt{5}-1\right)\)
\(=\sqrt{9-5}\times\sqrt{6+2\sqrt{5}}\left(\sqrt{5}-1\right)\)
\(=2\sqrt{\left(\sqrt{5}+1\right)^2}\left(\sqrt{5}-1\right)\)
\(=2\left(5-1\right)\)
= 8
a) \(A=\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
\(=\sqrt{5}-\sqrt{3-\sqrt{\left(3-2\sqrt{5}\right)^2}}\)
\(=\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}\)
\(=\sqrt{5}-\sqrt{3-2\sqrt{5}+3}\)
\(=\sqrt{5}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(=\sqrt{5}-\left(\sqrt{5}-1\right)\)
\(=\sqrt{5}-\sqrt{5}+1\)
\(=1\)
b) \(B=\sqrt{3-\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{3-\sqrt{5}}\left(3\sqrt{10}+\sqrt{50}-3\sqrt{2}-\sqrt{10}\right)\)
\(=\sqrt{3-\sqrt{5}}\left(3\sqrt{10}+5\sqrt{2}-3\sqrt{2}-\sqrt{10}\right)\)
\(=\sqrt{3-\sqrt{5}}\left(2\sqrt{10}+2\sqrt{2}\right)\)
\(=\sqrt{3-\sqrt{5}}\sqrt{\left(2\sqrt{10}+2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{5}\right)\left(2\sqrt{10}+2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{5}\right)\left(4\cdot10+8\sqrt{20}+4\cdot2\right)}\)
\(=\sqrt{\left(3-\sqrt{5}\right)\left(40+16\sqrt{5}+8\right)}\)
\(=\sqrt{\left(3-\sqrt{5}\right)\left(48+16\sqrt{5}\right)}\)
\(=\sqrt{\left(3-\sqrt{5}\right)\cdot16\left(3+\sqrt{5}\right)}\)
\(=\sqrt{\left(9-5\right)\cdot16}\)
\(=\sqrt{4\cdot16}\)
\(=\sqrt{64}\)
\(=8\)
\(B=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\sqrt{2}.\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{2}.\sqrt{3-\sqrt{5}}\)
\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\sqrt{2}.\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{6-2\sqrt{5}}\)
\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\sqrt{\left(\sqrt{5}+1\right)^2}+\left(3+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-1^2\right)}\)
\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left|\sqrt{5}-1\right|\)
\(=3\sqrt{5}+3-5-\sqrt{5}+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(=6\sqrt{5}-2\sqrt{5}=4\sqrt{5}\)
\(\Rightarrow B=\frac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)
Đặt \(\sqrt{3+\sqrt{5}}=a>0;\sqrt{3-\sqrt{5}}=b>0\Rightarrow ab=\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\sqrt{3^2-5}=2\)
Và \(a^2+b^2=6 \Rightarrow\left(a+b\right)^2=a^2+b^2+2ab=6+4=10\Rightarrow a+b=\sqrt{10}\) (vì a + b > 0 do a > 0,b>0)
\(B=b^2\cdot a+a^2\cdot b=ab\left(a+b\right)=2\sqrt{10}\)
\(=\sqrt{3-\sqrt{5}}.\sqrt{2}.\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)\)
\(=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=2\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=2\left(9-\left(\sqrt{5}\right)^2\right)\)
\(=2.4=8\)
Chỉ vậy thôi nha bạn ^_^
\(C=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}.\sqrt{3+\sqrt{5}.}\sqrt{2}\left(\sqrt{5}-1\right)\)
\(C=\sqrt{4}.\sqrt{6+2\sqrt{5}}\left(\sqrt{5}-1\right)\)
\(C=2.\sqrt{\left(\sqrt{5}+1\right)^2}.\left(\sqrt{5}-1\right)\)
\(C=2.\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=2.4=8\)