\(B=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\)

    \(=\left(2\sqrt{3}-1\right)^2\left(2+\sqrt{3}\right)^2-8\sqrt{20+2\sqrt{\left(4+3\sqrt{3}\right)^2}}\)

    \(=\left(3\sqrt{3}+4\right)^2-8\sqrt{20+2\left(4+3\sqrt{3}\right)}\)

    \(=\left(3\sqrt{3}+4\right)^2-8\sqrt{28+6\sqrt{3}}\)

    \(=\left(3\sqrt{3}+4\right)^2-8\sqrt{\left(3\sqrt{3}+1\right)^2}\)

    \(=43+24\sqrt{3}-8\left(3\sqrt{3}+1\right)=35\)

\(D=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\)

   \(=\left(2\sqrt{3}-1\right)^2\left(\sqrt{3}+2\right)^2-8\sqrt{20+2\left(3\sqrt{3}+4\right)}\)

   \(=\left(4+3\sqrt{3}\right)^2-8\sqrt{28+6\sqrt{3}}\)\(=\left(4+3\sqrt{3}\right)^2-8\left(3\sqrt{3}+1\right)\)

   \(=43+24\sqrt{3}-24\sqrt{3}-8=35\)

TÁch nó theo hằng đẳng thức ấy

Nhờ tách hộ cái.   Không biết làm mới lên đây hỏi

dùng a²+2ab+b²=(a+b)²

\(A=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-4}+\frac{\sqrt{x}-10}{x-4}\)

\(A=\frac{x+2\sqrt{x}+x-3\sqrt{x}+2+\sqrt{x}-10}{x-4}\)

\(A=\frac{2x-8}{x-4}=\frac{2\left(x-4\right)}{x-4}=2\)

\(B=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{\left(3\sqrt{3}+4\right)^2}}\)

\(B=43+24\sqrt{3}-8\sqrt{20+6\sqrt{3}+8}\)

\(B=43+24\sqrt{3}-8\sqrt{28+6\sqrt{3}}\)

\(B=43+24\sqrt{3}-8\sqrt{\left(3\sqrt{3}+1\right)^2}\)

\(B=43+24\sqrt{3}-24\sqrt{3}-8\)

\(B=35\)

Nguyễn Việt Lâm giúp mk nhá, tks bn nhìu :>>

\(A=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\)   \(A=\left(6+7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{\left(16+2.4.3\sqrt{3}+27\right)}}\)

\(A=6\left(7+4\sqrt{3}\right)+\left(7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{\left(4+3\sqrt{3}\right)^2}}\)Trong căn là hằng đẳng thức (a+b)^2

\(A=42+24\sqrt{3}+7^2-\left(4\sqrt{3}\right)^2-8\sqrt{20+2\left(4+3\sqrt{3}\right)}\) sử dụng hằng đẳng thức a^2 -b^2\(A=43+24\sqrt{3}-8\sqrt{20+8+2.3\sqrt{3}}\)

\(A=43+24\sqrt{3}-8\sqrt{1+2.3\sqrt{3}+27}\)trong căn tiếp tục là hằng đẳng thức (a+b)^2\(A=43+24\sqrt{3}-8\sqrt{\left(1+3\sqrt{3}\right)^2}\)

\(A=43+24\sqrt{3}-8\left(1+3\sqrt{3}\right)\)

\(A=35\)

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\(A=43+24\sqrt{3}-8\sqrt{20+2\sqrt{\left(3\sqrt{3}+4\right)^2}}\)

\(=43+24\sqrt{3}-8\sqrt{20+2\left(3\sqrt{3}+4\right)}\)

\(=43+24\sqrt{3}-8\sqrt{28+6\sqrt{3}}\)

\(=43+24\sqrt{3}-8\sqrt{\left(3\sqrt{3}+1\right)^2}\)

\(=43+24\sqrt{3}-8\left(3\sqrt{3}+1\right)\)

\(=43-8=35\)