2,6g dung dịch 

ez

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

\(=\frac{4}{2}=2\)

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

\(=\frac{4+4\sqrt{2}+2+4-4\sqrt{2}+2}{4-2}\)

\(=\frac{8+4}{2}=\frac{12}{2}=6\)

GIÚP MK NHA CÁC BN

\(\frac{\sqrt{8-\sqrt{15}}}{\sqrt{30}-\sqrt{2}}=\frac{\sqrt{2}\sqrt{8-\sqrt{15}}}{\sqrt{2}\left(\sqrt{15}.\sqrt{2}-\sqrt{2}\right)}=\frac{\sqrt{16-2\sqrt{15}}}{\sqrt{2}.\sqrt{2}\left(\sqrt{15}-1\right)}\)

\(=\frac{\sqrt{15-2\sqrt{15}+1}}{2\left(\sqrt{15}-1\right)}=\frac{\sqrt{\left(\sqrt{15}-1\right)^2}}{2\left(\sqrt{15}-1\right)}=\frac{\sqrt{15}-1}{2\left(\sqrt{15}-1\right)}=\frac{1}{2}\)

Đặt: \(x^2=t\)

Sao đó giải như pt bậc 2 bình thường 

cops mạng đâu thế :((

\(\Leftrightarrow x^2+\frac{x^2}{\left(x+1\right)^2}-2\frac{x^2}{x+1}+2\frac{x^2}{x+1}-1=0\)

\(\Leftrightarrow\left(x-\frac{x}{x+1}\right)^2+2\frac{x^2}{x+1}-1=0\)

\(\Leftrightarrow\left(\frac{x^2}{x+1}\right)^2+2\frac{x^2}{x+1}-1=0\)

Ta có: \(2021^2=\left(2020+1\right)^2=2020^2+2.2020.1+1^2\)

\(\Rightarrow1+2020^2=2021^2-2.2020\)

\(\Rightarrow\sqrt{1+2020^2+\frac{2020^2}{2021}}+\frac{2020}{2021}\)

\(=\sqrt{2021^2-2.2020+\frac{2020^2}{2021}}+\frac{2020}{2021}\)

\(=\sqrt{2021^2-2.2021.\frac{2020}{2021}+\left(\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)

\(=\sqrt{\left(2021-\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)

\(=2021-\frac{2020}{2021}+\frac{2020}{2021}=2021\)