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

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

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

\(\Rightarrow\frac{3\sqrt{3}-5}{3\sqrt{5}-5}=1\left(đpcm\right)\)

\(a,\sqrt{2x+5}=\sqrt{1-x}\)

\(\Rightarrow2x+5=1-x\)

\(2x+x=1-5\)

\(3x=-4\Leftrightarrow x=\frac{-4}{3}\)

Vậy \(S=\left\{-\frac{4}{3}\right\}\)thuộc tập nghiệm của pt trên

Bài 1:

\(\sqrt{24+8\sqrt{15}-\sqrt{9-4\sqrt{5}}}\)

\(=\sqrt{24+8\sqrt{15}-\left(\sqrt{5}-2\right)}\)

\(=\sqrt{26+8\sqrt{15}-\sqrt{5}}\)

Bài 2:

\(A=\sqrt{\frac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}\)

\(A=\sqrt{\frac{x^4+6x^2+9}{x^2}}\)

\(A=\frac{\sqrt{x^4+6x^2+9}}{\sqrt{x^2}}\)

\(A=\frac{\sqrt{\left(x^2+3\right)^2}}{x}\)

\(A=\frac{x^2+3}{x}\)

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

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

wrecking ball sai rồi \(\frac{\sqrt{\left(x^2+3\right)^2}}{x}=\frac{trituyetdoix^2+3}{x}\) bằng 

Ta có:  \(\left(x+\sqrt{x^2+3}\right)\left(\sqrt{x^2+3}-x\right)=3\)

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

Kết hợp với giả thiết ta có:

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

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

Cộng theo vế ta được: \(-\left(x+y\right)=x+y\)

\(\Rightarrow\)\(E=x+y=0\)

\(\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\)

\(\Leftrightarrow\left(x+\sqrt{x^2+3}\right)\left(x-\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\left(x-\sqrt{x^2+3}\right)\)

\(\Leftrightarrow\left(x^2-x^2-3\right)\left(y+\sqrt{y^2+3}\right)=3\left(x-\sqrt{x^2+3}\right)\)

\(\Leftrightarrow-3\left(y+\sqrt{y^2+3}\right)=3\left(x-\sqrt{x^2+3}\right)\)

\(\Leftrightarrow-y-\sqrt{y^2+3}=x-\sqrt{x^2+3}\)(*)

Tương tự, nhân mỗi vế vs \(y-\sqrt{y^2+3}\), ta được:

\(-x-\sqrt{x^2+3}=y-\sqrt{y^2+3}\)(**)

Cộng (*) và (**) suy ra :

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

\(\Leftrightarrow-y-x=x+y\Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\)

Vậy \(E=0.\)