\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\\ =\sqrt{2}.\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{2}.\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\\ =\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{4+2\sqrt{3}}}+\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{4-2\sqrt{3}}}\\ =\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\dfrac{\left|\sqrt{3}-1\right|}{\left|\sqrt{3}+1\right|}+\dfrac{\left|\sqrt{3}+1\right|}{\left|\sqrt{3}-1\right|}\\ =\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\\ =\dfrac{4-2\sqrt{3}+4+2\sqrt{3}}{\sqrt{3^2}-1}\\ =\dfrac{8}{2}\\ =4\)

A = \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\) + \(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)  = \(\dfrac{\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}}{\sqrt{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}}\)

A = \(\dfrac{2-\sqrt{3}+2+\sqrt{3}}{\sqrt{4-3}}\) = \(\dfrac{4}{1}\) = 4

Lời giải:

$A=x^2-2mx+m+1=(x-m)^2+m+1-m^2\geq m+1-m^2$

$A_{\min}=m+1-m^2=11$

$\Leftrightarrow m^2-m+10=0$

$\Leftrightarrow (m-\frac{1}{2})^2=\frac{-39}{4}<0$ (vô lý)

Vậy hông tồn tại $m$ để $A_{\min}=11$

\(\dfrac{2-\sqrt{5}}{2+\sqrt{5}}+\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\)

\(=\dfrac{\left(2-\sqrt{5}\right)^2}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}+\dfrac{\left(\sqrt{5}+2\right)^2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)

\(=\dfrac{4-4\sqrt{5}+5}{4-5}+\dfrac{5+4\sqrt{5}+4}{5-4}\)

\(=-4+4\sqrt{5}-5+5+4\sqrt{5}+4\)

\(=8\sqrt{5}\)

\(=\dfrac{\left(2-\sqrt{5}\right)\left(2-\sqrt{5}\right)}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}+\dfrac{\left(\sqrt{5}+2\right)^2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)

\(=\dfrac{9-4\sqrt{5}}{-1}+9+4\sqrt{5}\)

=9+4căn 5-9+4căn 5

=8*căn 5

Lời giải:
Áp dụng BĐT Bunhiacopxky:

$(a^2+b^2+c^2)(1+1+1)\geq (a+b+c)^2$

$\Rightarrow a^2+b^2+c^2\geq \frac{(a+b+c)^2}{3}$

$\Rightarrow (a^2+b^2+c^2)^3\geq \frac{(a+b+c)^6}{27}$

Áp dụng BĐT Cô-si: $a+b+c\geq 3\sqrt[3]{abc}=3$

$\Rightarrow (a^2+b^2+c^2)^3\geq \frac{(a+b+c)^6}{27}\geq \frac{(a+b+c).3^5}{27}=9(a+b+c)$
Ta có đpcm

Dấu "=" xảy ra khi $a=b=c=1$

\(\dfrac{1}{1-\sqrt{2}}+\dfrac{1}{1+\sqrt{2}}\)

\(=\dfrac{1+\sqrt{2}}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}+\dfrac{1-\sqrt{2}}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}\)

\(=\dfrac{1+\sqrt{2}+1-\sqrt{2}}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}\)

\(=\dfrac{2}{1-2}\)

\(=-2\)

\(=\dfrac{1+\sqrt{2}+1-\sqrt{2}}{1-2}=\dfrac{2}{-1}=-2\)

\(=\left(x\sqrt{x}+y\sqrt{y}\right)+\left(x-y\right)\)

\(=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\left(\sqrt{x}+\sqrt{y}\right)\left(x+y-\sqrt{xy}+\sqrt{x}-\sqrt{y}\right)\)

\(\dfrac{3\sqrt{5}}{2\sqrt{5}-1}\)

\(=\dfrac{3\sqrt{5}\left(2\sqrt{5}+1\right)}{\left(2\sqrt{5}-1\right)\left(2\sqrt{5}+1\right)}\)

\(=\dfrac{30-3\sqrt{5}}{\left(2\sqrt{5}\right)^2-1}\)

\(=\dfrac{30-\sqrt{5}}{20-1}\)

\(=\dfrac{30-\sqrt{5}}{19}\)

\(\dfrac{2-\sqrt{6}}{3\sqrt{3}-3}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)}{3\left(\sqrt{3}-1\right)}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{3}+1\right)}{3\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{6}+\sqrt{2}-3-\sqrt{3}\right)}{3\left(3-1\right)}\)

\(=\dfrac{2\sqrt{3}+2-3\sqrt{2}-\sqrt{6}}{6}\)

\(=\dfrac{\left(2-\sqrt{6}\right)\left(\sqrt{3}+1\right)}{3\cdot2}=\dfrac{2\sqrt{3}+2-3\sqrt{2}-\sqrt{6}}{6}\)