Ta có : \(\frac{23\sqrt{2}}{\sqrt{2}+\sqrt{14+5\sqrt{3}}}=\frac{46}{2+\sqrt{28+10\sqrt{3}}}=\frac{46}{2+\sqrt{\left(\sqrt{3}+5\right)^2}}=\frac{46}{7+\sqrt{3}}\)

\(=\frac{46\left(7-\sqrt{3}\right)}{\left(7+\sqrt{3}\right)\left(7-\sqrt{3}\right)}=\frac{46\left(7-\sqrt{3}\right)}{46}=7-\sqrt{3}\)

Suy ra a = 7 , b = -1

=> a x b = -7

\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\frac{6+2\sqrt{5}}{4}=\frac{32-6-2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}=\frac{14-\sqrt{5}}{2}\) \(\left(\frac{9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2-\left(\frac{9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2=\left(\frac{9-2\sqrt{14}-9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)\left(\frac{9-2\sqrt{14}+9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)=\frac{-72\sqrt{14}}{\sqrt{7}-\sqrt{2}}\)

Bài 1:
Xét tử số:

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

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

Xét mẫu số:
\(\sqrt{(\sqrt{5}+1)\sqrt{6-2\sqrt{5}}}=\sqrt{(\sqrt{5}+1)\sqrt{5+1-2\sqrt{5}}}=\sqrt{(\sqrt{5}+1)\sqrt{(\sqrt{5}-1)^2}}\)

\(=\sqrt{(\sqrt{5}+1)(\sqrt{5}-1)}=\sqrt{4}=2\)

Do đó: $A=\frac{2\sqrt{5}}{2}=\sqrt{5}$

Bài 2:

a)

$B=(\sqrt[3]{2}+1)^3(\sqrt[3]{2}-1)^3$
$=[(\sqrt[3]{2}+1)(\sqrt[3]{2}-1)]^3$
$=(\sqrt[3]{4}-1)^3$

$=3-3\sqrt[3]{16}+3\sqrt[3]{4}$

b)

Với $a,b$ đã cho ta đặt $\sqrt[3]{2}=x$. Khi đó:

\(a=\frac{6}{2x-2+\frac{2}{x}}=\frac{3x}{x^2-x+1}=\frac{3x(x+1)}{x^3+1}=\frac{3x(x+1)}{2+1}=x(x+1)\)

\(b=\frac{2}{2x+2+\frac{2}{x}}=\frac{x}{x^2+x+1}=\frac{x(x-1)}{x^3-1}=\frac{x(x-1)}{2-1}=x(x-1)\)

Khi đó:

$C=a^3b-ab^3=ab(a^2-b^2)=ab(a-b)(a+b)$

$=x^2(x^2-1)(2x)(2x^2)=4x^5(x^2-1)=8\sqrt[3]{4}(\sqrt[3]{4}-1)$

Bài 3:

Ta biết rằng $x^2-x+1=(x-\frac{1}{2})^2+\frac{3}{4}>0$ với mọi $x\in\mathbb{R}$

Do đó:

$|x^2-x+1|-|x-2|=6$

$\Leftrightarrow x^2-x+1-|x-2|=6(*)$

Nếu $x\geq 2$ thì $(*)\Leftrightarrow x^2-x+1-(x-2)=6$

$\Leftrightarrow x^2-2x-3=0$

$\Leftrightarrow (x-3)(x+1)=0$

$\Leftrightarrow x=3$ (do $x\geq 2$)

Nếu $x< 2$ thì $(*)\Leftrightarrow x^2-x+1-(2-x)=6$

$\Leftrightarrow x^2-7=0$

$\Rightarrow x=-\sqrt{7}$ (do $x< 2$)

Vậy........

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

Biến đổi vế trái :

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

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

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

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

\(=\frac{\sqrt{2}\left(6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3\right)}{9-3}=\frac{6\sqrt{2}}{6}=\sqrt{2}=VP\left(đpcm\right)\)

b) \(B=\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=8\)

Biến đổi vế trái :

VT = \(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\sqrt{5+\sqrt{21}}\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5+\sqrt{21}}\sqrt{5-\sqrt{21}}\)

\(=\sqrt{2}\sqrt{5+\sqrt{21}}\left(\sqrt{7}-\sqrt{3}\right)\sqrt{25-21}=\sqrt{10+2\sqrt{21}}\left(\sqrt{7}-\sqrt{3}\right)\sqrt{4}=\left|\sqrt{7}+\sqrt{3}\right|\left(\sqrt{7}-\sqrt{3}\right)2\)

\(=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)2=\left(7-3\right)2=4.2=8=VP\left(đpcm\right)\)

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

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

\(=\frac{6\sqrt{2}-2\sqrt{6}+3\sqrt{6}-\sqrt{18}+6\sqrt{2}+2\sqrt{6}-3\sqrt{6}-\sqrt{18}}{6}\)

\(=\frac{12\sqrt{2}-2\sqrt{18}}{6}=\frac{6\sqrt{2}}{6}=\sqrt{2}\)