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Làm luôn nhé
\(2B=21.2\left[\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)\right]^2-2.15\sqrt{15}\)
\(2B=21\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-6\left(\sqrt{3}-1+\sqrt{5}-1\right)^2-30\sqrt{15}\)
\(2B=21\left(\sqrt{3}+\sqrt{5}\right)^2-6\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(8+2\sqrt{15}\right)-30\sqrt{15}\)
\(2B=120+30\sqrt{15}-30\sqrt{5}\)
\(2B=120\)
\(B=60\)
\(c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
\(=\sqrt{4+5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{29}\)
a) \(\dfrac{2\sqrt{125}-3\sqrt{5}-\sqrt{180}}{-\sqrt{5}}+\sqrt{8}=\dfrac{2\sqrt{25.5}-3\sqrt{5}-\sqrt{36.5}}{-\sqrt{5}}+\sqrt{8}\)
\(=\dfrac{10\sqrt{5}-3\sqrt{5}-6\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=\dfrac{\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=2\sqrt{2}-1\)
b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}\)
\(=\sqrt{3}-\sqrt{2}+3\sqrt{2}=2\sqrt{2}+\sqrt{3}\)
c) \(\sqrt{48}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}=\sqrt{16.3}-2\sqrt{9.\dfrac{1}{3}}+\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}\)
\(=4\sqrt{3}-2\sqrt{3}+1-\sqrt{3}=1+\sqrt{3}\)
d) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3\)
1. \(=\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{3}{2}}\right)^2}+\sqrt{\left(\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}\right)^2}-2\sqrt{4\sqrt{7}}=\frac{7}{2}+\frac{3}{2}+\frac{7}{2}-\frac{3}{2}-2\sqrt{4\sqrt{7}}\)
\(=7-2\sqrt{4\sqrt{7}}\)
cho hỏi tại sao có số \(\frac{7}{2};\frac{3}{2}\)zậy chỉ với
a: \(=\left(\sqrt{3}-2\right)\cdot\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\)
=3-4=-1
b: \(=\sqrt{6+4\sqrt{2}}-\sqrt{11-2\sqrt{18}}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=2+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}-1\)
c: \(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)
\(=2\sqrt{5}-1+2\sqrt{5}+1\)
\(=4\sqrt{5}\)
a, \(=>3-\sqrt{2}+\sqrt{50}=3-\sqrt{2}+5\sqrt{2}=3+4\sqrt{2}\)
b, \(=>\dfrac{\sqrt[3]{125.5}}{\sqrt[3]{5}}-\sqrt[3]{\left(-4\right).2}=\sqrt[3]{125}-\sqrt[3]{\left(-2\right)^3}\)
\(=5-\left(-2\right)=7\)
c, \(=>\sqrt{6}.\sqrt{\dfrac{6}{2}}-\sqrt{2}-3\sqrt{4.2}=\sqrt{6}.\sqrt{3}-\sqrt{2}-6\sqrt{2}\)
\(=\sqrt{18}-7\sqrt{2}=3\sqrt{2}-7\sqrt{2}=-4\sqrt{2}\)
d, \(=>\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-\dfrac{2}{\sqrt{3}-1}=\sqrt{3}-\dfrac{2}{\sqrt{3}-1}\)
\(=\dfrac{3-\sqrt{3}-2}{\sqrt{3}-1}=\dfrac{1-\sqrt{3}}{\sqrt{3}-1}=-1\)
\(A=\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)
\(A=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)
\(A=2^2-\left(\sqrt{5}\right)^2\)
\(A=4-5\)
\(A=-1\)
____
\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(B=6-121\)
\(B=-115\)
