\(5.\dfrac{5}{\sqrt{21}-4}+\dfrac{3\sqrt{7}-7\sqrt{3}}{\sqrt{7}-\sqrt{3}}\\ =\dfrac{5\left(\sqrt{21}+4\right)}{21-16}+\dfrac{\sqrt{3}.\sqrt{7}\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{7}-\sqrt{3}}\\ =\dfrac{5\left(\sqrt{21}+4\right)}{5}+\sqrt{21}\\ =\sqrt{21}+4+\sqrt{21}=4+2\sqrt{21}\)

\(6.\dfrac{8}{3+\sqrt{5}}-\dfrac{\sqrt{15}-2\sqrt{5}}{\sqrt{3}-2}\\ =\dfrac{8\left(3-\sqrt{5}\right)}{9-5}-\dfrac{\left(\sqrt{15}-2\sqrt{5}\right)\left(\sqrt{3}+2\right)}{3-4}\\ =2\left(3-\sqrt{5}\right)+\sqrt{15}.\sqrt{3}+2\sqrt{15}-2\sqrt{15}-4\sqrt{5}\\ =6-2\sqrt{5}+3\sqrt{5}+2\sqrt{15}-2\sqrt{15}-4\sqrt{5}\\ =6-3\sqrt{5}\)

\(b)x^2-5x-2(x-5)=0\\<=>(x^2-5x)-2(x-5)=0\\<=>x(x-5)-2(x-5)=0\\<=>(x-5)(x-2)=0\\\)

\(< =>\left[{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

Vậy x =5 hoặc x =2

Bài 4:

)\(\left[\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right]\dfrac{x-4}{\sqrt{4x}}\\ DKXD:x>4;x\ne0\\ =\left[\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}\right]\dfrac{x-4}{2\sqrt{x}}\\ =\left[\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\right]\dfrac{x-4}{2\sqrt{x}}\\ =\dfrac{2x}{x-4}.\dfrac{x-4}{2\sqrt{x}}=\sqrt{x}\\ b.A=3\\ \Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\left(tmdk\right)\)

Vậy x=9 thì A=3

Bài 1:

\(a.7\sqrt{24}-\sqrt{150}=7\sqrt{4.6}-\sqrt{25.6}\\ =14\sqrt{6}-5\sqrt{6}=7\sqrt{6}\\ b.\left(\sqrt{12}-\sqrt{48}-\sqrt{108}-\sqrt{192}\right):2\sqrt{3}\\ =\left(2\sqrt{3}-4\sqrt{3}-6\sqrt{3}-8\sqrt{3}\right):2\sqrt{3}\\ =-\dfrac{16\sqrt{3}}{2\sqrt{3}}=-8\\ 2.\sqrt{2x-3}=5\\ DKXD:x\ge\dfrac{3}{2}\\ < =>2x-3=5^2\\ < =>2x-3=25\\ < =>2x=28\\ < =>x=14\left(tmdk\right)\\ =>x=14\)

Bài 3:

\(a.7\sqrt{2}=\sqrt{7^2.2}=\sqrt{98}\)

Vì \(\sqrt{98}>\sqrt{72}\Rightarrow7\sqrt{2}>\sqrt{72}\)

\(b.4\sqrt[]{3}=\sqrt{4^2.3}=\sqrt{48}\\ 3\sqrt{5}=\sqrt{3^2.5}=\sqrt{45}\)

vì: \(48>45\Rightarrow\sqrt{48}>\sqrt{45}\Rightarrow4\sqrt{3}>3\sqrt{5}\)

c.\(4\sqrt{7}=\sqrt{4^2.7}=\sqrt{112}\\ 5\sqrt{6}=\sqrt{5^2.6}=\sqrt{150}\)

Vì: \(112< 150\Rightarrow\sqrt{112}< \sqrt{150}\Rightarrow4\sqrt{7}< 5\sqrt{6}\)

d.\(\dfrac{1}{6}\sqrt{18}=\sqrt{\dfrac{18}{6^2}}=\sqrt{\dfrac{18}{36}}=\sqrt{\dfrac{1}{2}}\\ \dfrac{1}{2}\sqrt{2}=\sqrt{\dfrac{2}{4}}=\sqrt{\dfrac{1}{2}}\)

Vì: \(\dfrac{1}{2}=\dfrac{1}{2}\Rightarrow\sqrt{\dfrac{1}{2}}=\sqrt{\dfrac{1}{2}}\Rightarrow\dfrac{1}{6}\sqrt{15}=\dfrac{1}{2}\sqrt{2}\)

Bài 2:

\(a.\left(5-a\right)\sqrt{\dfrac{8a}{a-5}}\\ a>5 \\= \sqrt{\dfrac{8a\left(5-a\right)^2}{a-5}}\\ =\sqrt{\dfrac{8a\left(a-5\right)^2}{a-5}}\\ =\sqrt{8a\left(a-5\right)}=\sqrt{8a^2-40}\\b.\left(x-7\right)\sqrt{\dfrac{\left(x+7\right)}{49-x^2}}\\ =\left(x-7\right)\sqrt{\dfrac{\left(x+7\right)}{\left(7-x\right)\left(x+7\right)}}\\ \left(x-7\right)\sqrt{\dfrac{1}{7-x}}\\ =\sqrt{\dfrac{\left(x-7\right)^2}{7-x}}=\sqrt{\dfrac{\left(7-x\right)^2}{7-x}}\\ =\sqrt{7-x}\\ c.\)

\(\dfrac{a}{b}\sqrt{\dfrac{b}{a}}\\ a,b>0 \\ =\sqrt{\dfrac{a^2b}{b^2a}}=\sqrt{\dfrac{a}{b}}\\ d.x\sqrt{\dfrac{6}{x}}\\ x>0\\ =\sqrt{\dfrac{6x^2}{x}}=\sqrt{6x}\)