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rút gọn biểu thức
\(D=\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}}\)
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)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
a: \(=-6\sqrt{b}-\dfrac{1}{3}\cdot3\sqrt{3b}+\dfrac{1}{5}\cdot5\sqrt{6b}\)
\(=-6\sqrt{b}-\sqrt{3}\cdot\sqrt{b}+\sqrt{6}\cdot\sqrt{b}\)
\(=\sqrt{b}\left(-6-\sqrt{3}+\sqrt{6}\right)\)
c: \(=\sqrt{\left(5+2\sqrt{6}\right)^2}+\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=5+2\sqrt{6}+5-2\sqrt{6}=10\)
d: \(A=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
e: \(B=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}\)
\(=\sqrt{6+2\cdot\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
a) Đặt \(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}\)
\(A^2=5-2\sqrt{6}+2\sqrt{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}+5+2\sqrt{6}\)
\(=10+2\sqrt{25-4.6}=10+2\sqrt{1}=10+2=12\)
\(\Rightarrow A=\sqrt{12}\)
b)\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}.\sqrt{5}-\sqrt{2}}{\sqrt{5}-1}+\frac{\sqrt{2}.\sqrt{2}-\sqrt{2}}{\sqrt{2}-1}\)
\(=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
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) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)
\(\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}+\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}-\sqrt{5}}\)
\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)^2}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}+\frac{\left(\sqrt{6}+\sqrt{5}\right)^2}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)
\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)^2+\left(\sqrt{6}+\sqrt{5}\right)^2}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)
\(=\frac{11-2\sqrt{30}+11+2\sqrt{30}}{\left(\sqrt{6}\right)^2-\left(\sqrt{5}\right)^2}\)
\(=\frac{22}{1}=22\)
\(\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}+\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}-\sqrt{5}}\)
\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)+\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)
\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)^2+\left(\sqrt{6}+\sqrt{5}\right)^2}{\sqrt{6}^2+\sqrt{5}^2}\)
\(=\sqrt{6}^2-2\sqrt{6}.\sqrt{5}+\sqrt{5}^2+\sqrt{6}^2+2\sqrt{6}.\sqrt{5}+\sqrt{5}^2\)
\(=6+5+6+5=22\)
\(\sqrt{14+6\sqrt{5}}-\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}}=5\)
\(\sqrt{14+6\sqrt{5}}-\sqrt{\frac{\sqrt{5-2}}{\sqrt{5}+2}}=5\)
*****~~~~~~~~~~*****
\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{6+\sqrt{6}}{\sqrt{6}+1}\)
\(=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{\sqrt{6}\left(\sqrt{6}+1\right)}{\sqrt{6}+1}\)
\(=\sqrt{3}+\sqrt{6}\)
\(=\sqrt{3}\left(1+\sqrt{2}\right)\)
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\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
\(=\sqrt{3}+2+\sqrt{2}\)
(Chúc bạn học tốt nha!)
\(D=\sqrt{\frac{\left(5+2\sqrt{6}\right)^2}{25-24}}+\sqrt{\frac{\left(5-2\sqrt{6}\right)^2}{25-24}}=5+2\sqrt{6}+5-2\sqrt{6}=10\)