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

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

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

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

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

\(=43-8=35\)

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

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

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

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

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

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

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

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

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

   \(=43+24\sqrt{3}-24\sqrt{3}-8=35\)

Câu 1: 

a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\) 

\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)

\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)

hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)

Câu 1: 

a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\) 

\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)

\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)

hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)​

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: Ta có: \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}-\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\dfrac{8}{8+2\sqrt{15}}-\dfrac{8}{8-2\sqrt{15}}\)

\(=\dfrac{64-16\sqrt{15}-64-16\sqrt{15}}{4}\)

\(=\dfrac{-32\sqrt{15}}{4}=-8\sqrt{15}\)

b: Ta có: \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)

\(=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{-2}\)

\(=-\dfrac{6\sqrt{2}}{2}=-3\sqrt{2}\)

b) \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{6\sqrt{2}}{-2}=-3\sqrt{2}\)

c) \(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right):\sqrt{28}=\dfrac{\left(\sqrt{7}+3\right)^2-\left(\sqrt{7}-3\right)^2}{\left(\sqrt{7}-3\right)\left(\sqrt{7}+3\right)}:\sqrt{28}=\dfrac{16+6\sqrt{7}-16+6\sqrt{7}}{7-9}=\dfrac{12\sqrt{7}}{-2}=-6\sqrt{7}\)