1.

\(A=2\sqrt{12}-\sqrt{48}+3\sqrt{\dfrac{4}{3}}.\\ A=4\sqrt{3}-4\sqrt{3}+2\sqrt{3}.\\ A=0+2\sqrt{3}=2\sqrt{3}.\\ B=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\left(x>0;x\ne1\right).\\ B=\dfrac{x}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}.\\ B=\dfrac{x}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}.\\ B=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}.\\ B=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}.\\ B=\sqrt{x}-1.\)

2.

Thay \(x=6+2\sqrt{5}\) (TMĐK) vào biểu thức B, ta có:

\(B=\sqrt{6+2\sqrt{5}}-1.\)

\(\left(a+b\right)^3-3ab\left(a+b\right).\\ =a^3+3a^2b+3ab^2+b^3-3a^2b+3ab^2.\\ =a^3+b^3.\)

Xét \(\Delta ABC:\widehat{BAC}+\widehat{B}+\widehat{C}=180^o\) (Tổng 3 góc trong tam giác).

Mà \(\widehat{B}=\widehat{C}=50^o\left(gt\right).\)

\(\Rightarrow\widehat{BAC}=80^o.\)

\(\widehat{BAC}+2\widehat{BAx}=180^o.\\ \Rightarrow80^o+2\widehat{BAx}=180^o.\\ \Rightarrow\widehat{BAx}=50^o.\)

Có: \(\widehat{B}=50^o\left(gt\right).\)

\(\Rightarrow\widehat{B}=\widehat{BAx}.\)

Mà 2 góc này ở vị trí so le trong.

\(\Rightarrow Ax//BC\left(dhnb\right).\)