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a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)
b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)
a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)
\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(2-5\right)\)
\(=-\left(-3\right)\)
\(=3\)
b) Ta có:
\(x^2-x\sqrt{3}+1\)
\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)
a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
B=\(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}=\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
C=\(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=\frac{3\left(1+\sqrt{3}\right)}{\sqrt{3}}+\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)
D=\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
E=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)
a ) \(A=\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{\left(\sqrt{5}-\sqrt{3}\right)-\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}\)
\(=\frac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{5-3}\)
\(=\frac{-2\sqrt{3}}{2}\)
\(=-\sqrt{3}\)
c ) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
\(=\frac{1}{2+\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{2}{\sqrt{3}\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+1\right)+\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)-2\left(2+\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{2\sqrt{3}+4}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{2\left(\sqrt{3}+2\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{2.\sqrt{3}\left(\sqrt{3}-1\right)}{3\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3.\left(3-1\right)}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{3}\)
\(=\frac{3-\sqrt{3}}{3}\)
\(=1-\frac{\sqrt{3}}{3}\)
a) \(A=\frac{3-\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}}=\sqrt{3}-1\)
b) \(B=\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{5}+1}=\frac{\sqrt{5}+1}{\sqrt{5}+1}=1\)
c) \(C=\frac{2\sqrt{2}+\sqrt{6}}{4+\sqrt{12}}=\frac{2\sqrt{2}+\sqrt{6}}{4+2\sqrt{3}}=\frac{\left(2\sqrt{2}+\sqrt{6}\right)\left(4-2\sqrt{3}\right)}{\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)}=\frac{2\sqrt{2}}{4}=\frac{\sqrt{2}}{2}\)
d) \(D=\frac{\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{5+2\sqrt{6}}\left(\sqrt{2}-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=-\sqrt{5+2\sqrt{6}}\left(\sqrt{2}-\sqrt{3}\right)\)
a) \(\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{\left(2+\sqrt{3}\right)^2}{4-3}\)
\(=\left(2+\sqrt{3}\right)^2=7+4\sqrt{3}\)
\(\frac{5+2\sqrt{6}}{5-2\sqrt{6}}=\frac{\left(5+2\sqrt{6}\right)^2}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}=\frac{\left(5+2\sqrt{6}\right)^2}{25-24}\)
\(=\left(5+2\sqrt{6}\right)^2=49+20\sqrt{6}\)
b) \(\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{3-2\sqrt{3}+1}{3-1}\)
\(=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
c) \(\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2+\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}=14\)
d) \(\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}-\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}\)
\(=\frac{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2-\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{2+\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}-\left(2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\right)}{2+\sqrt{3}-\left(2-\sqrt{3}\right)}\)
\(=\frac{4\sqrt{4-3}}{2\sqrt{3}}=\frac{4}{2\sqrt{3}}=\frac{2}{\sqrt{3}}\)