\(=\left(\sqrt{3}+1\right)\sqrt{\frac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}}\)

\(=\left(\sqrt{3}+1\right)\sqrt{\frac{70-14\sqrt{3}-30\sqrt{3}+18}{25-\sqrt{3}^2}}\)

\(=\left(\sqrt{3}+1\right)\sqrt{\frac{88-44\sqrt{3}}{22}}\)

\(=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)

\(A=\frac{\left(\sqrt{3}+1\right)\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}=\frac{4\left(11\sqrt{3}-11\right)}{25^2-\left(\sqrt{3}\right)^2}=\frac{44\left(\sqrt{3}-1\right)}{22}=2\sqrt{3}-2\)

\(\frac{4\left(11\sqrt{3}-11\right)}{25^2-\left(\sqrt{3}\right)^2}\) Thay Bằng:

\(\frac{4\left(11\sqrt{3}-11\right)}{5^2-\left(\sqrt{3}\right)^2}\)

Cảm ơn bạn! 

tại mình ghi lộn á, 25 chứ k phải 25 bình đâu

mình không viết lại đề nha 

\(=\sqrt{\frac{\left(\sqrt{3}+1\right)^2.\left(14-6\sqrt{3}\right)}{5+\sqrt{3}}}\)

\(=\sqrt{\frac{\left(3+2\sqrt{3}+1\right).\left(14-6\sqrt{3}\right)}{5+\sqrt{3}}}\)

\(=\sqrt{\frac{\left(4+2\sqrt{3}\right).\left(14-6\sqrt{3}\right)}{5+\sqrt{3}}}\)

\(=\sqrt{\frac{56-24\sqrt{3}+28\sqrt{3}-36}{5+\sqrt{3}}}\)

\(=\sqrt{\frac{20+4\sqrt{3}}{5+\sqrt{3}}}\)

\(=\sqrt{\frac{\left(20+4\sqrt{3}\right).\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right).\left(5-\sqrt{3}\right)}}\)

\(=\sqrt{\frac{100-20\sqrt{3}+20\sqrt{3}-12}{5^2-\sqrt{3}^2}}\)

\(=\sqrt{\frac{88}{25-3}}\)

\(=\sqrt{\frac{88}{22}}\)

\(=\sqrt{4}\)

\(=2\)

HỌC TỐT !!! 

\(\sqrt{\left(\sqrt{5}+3\right)^2}+\sqrt{14-6\sqrt{5}}\)\(=\left|\sqrt{5}+3\right|+\sqrt{9-2.3\sqrt{5}+5}\)

\(=\sqrt{5}+3+\sqrt{\left(3-\sqrt{5}\right)^2}\) \(=\sqrt{5}+3+\left|3-\sqrt{5}\right|\)

\(=\sqrt{5}+3+3-\sqrt{5}=6\) ( do \(3-\sqrt{5}>0\))

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)

\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)

Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé

a) \(\frac{3}{2+\sqrt{3}}+\frac{13}{4-\sqrt{3}}+\frac{6}{\sqrt{3}}\)

\(=\frac{3\left(2-\sqrt{3}\right)}{2^2-3}+\frac{13\left(4+\sqrt{3}\right)}{4^2-3}+\frac{6\sqrt{3}}{3}\)

\(=3\left(2-\sqrt{3}\right)+\left(4+\sqrt{3}\right)+2\sqrt{3}\)

\(=3.2+4=6+4=10\)

b) \(=\left[\frac{\left(\sqrt{14}-\sqrt{7}\right)\left(\sqrt{2}+1\right)}{2-1}+\frac{\left(\sqrt{15}-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{3-1}\right]:\frac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\) (nhân bung mấy cái trong ngoặc vuông ra, rút gọn)

c) Gợi ý: \(28-10\sqrt{3}=5^2-2.5.\sqrt{3}+\sqrt{3}=\left(5-\sqrt{3}\right)^2\)

d) \(=\frac{3\left(3-2\sqrt{3}\right)}{3^2-\left(2\sqrt{3}\right)^2}+\frac{3\left(3+2\sqrt{3}\right)}{3^2-\left(2\sqrt{3}\right)^2}=-6\)

e) Tự làm.

Cái câu c đánh nhầm:

\(=5^2-2.5.\sqrt{3}+3=\left(5-\sqrt{3}\right)^2\) nha!