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!

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)

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

những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé

\({\left[ {{{\left( {\frac{1}{3}} \right)}^2}} \right]^{\frac{1}{4}}}.{\left( {\sqrt 3 } \right)^5} = {\left( {\frac{1}{3}} \right)^{2.\frac{1}{4}}}.{\left( {{3^{\frac{1}{2}}}} \right)^5} = {\left( {{3^{ - 1}}} \right)^{\frac{1}{2}}}{.3^{\frac{1}{2}.5}} = {3^{ - \frac{1}{2}}}{.3^{\frac{5}{2}}} = {3^{ - \frac{1}{2} + \frac{5}{2}}} = {3^2} = 9\)

Chọn D.

a) Ta có: \(A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)

\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\left(\sqrt{9}-\sqrt{4}\right)\)

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

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

\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)(Vì \(\sqrt{5}>\sqrt{3}\))

\(=5-3-\sqrt{5}\)

\(=2-\sqrt{5}\)

b) Ta có: \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)

\(=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)\)

\(=\sqrt{3}+\sqrt{3}+\sqrt{6}\)

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

c) Ta có: \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)

\(=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}\)

\(=2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+\sqrt{1}\)

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

\(=2\sqrt{3}+2-\sqrt{3}+\frac{2}{3}\)(Vì \(2>\sqrt{3}\))

\(=\sqrt{3}+\frac{8}{3}\)

d) Ta có: \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)

\(=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)

\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|\)

\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))

\(=6-3\sqrt{3}\)

Bài 1:

a)  \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)

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

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

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

b)   \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)

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

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

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

c)  ĐK:  \(a\ge0;a\ne1\)

  \(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)

\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)

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

\(=1-a+a=1\)