bằng 8

nhanh nha mn

\(\frac{3}{\sqrt{3}-2}+\frac{2}{2+\sqrt{3}}=\frac{3\left(\sqrt{3}+2\right)+2\left(\sqrt{3}-2\right)}{\left(\sqrt{3}-2\right)\left(2+\sqrt{3}\right)}=\frac{3\sqrt{3}+6+2\sqrt{3}-4}{3-4}\)

\(=\frac{5\sqrt{3}+2}{-1}=-5\sqrt{3}-2\)

Ta có: \(\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}=\frac{2017-1}{\sqrt{2017}}+\frac{2016+1}{\sqrt{2016}}=\sqrt{2017}+\sqrt{2016}+\left(\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}\right)\)

Do \(\sqrt{2016}< \sqrt{2017}\Rightarrow\frac{1}{\sqrt{2016}}>\frac{1}{\sqrt{2017}}\Rightarrow\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}>0\)

=> \(\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}>\sqrt{2017}+\sqrt{2016}\)

Hay \(\sqrt{2016}+\sqrt{2017}< \frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}\)

Ta có : \(\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}\)

\(=\frac{2017-1}{\sqrt{2017}}+\frac{2016+1}{\sqrt{2016}}\)

\(=\sqrt{2016}+\sqrt{2017}+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}\)

Vì \(\sqrt{2016}< \sqrt{2017}\)\(\Rightarrow\frac{1}{\sqrt{2016}}>\frac{1}{\sqrt{2017}}\)

\(\Rightarrow\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}>0\)

\(\Rightarrow\sqrt{2016}+\sqrt{2017}+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}>\sqrt{2016}+\sqrt{2017}\)

\(\Rightarrowđpcm\)

Đk: \(-1\le a\le1\)

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

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

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

\(P=\frac{\sqrt{2+2\sqrt{1-a^2}}\left(\sqrt{1+a}-\sqrt{1-a}\right).\left(2+\sqrt{1-a^2}\right)}{2\left(2+\sqrt{1+a^2}\right)}\)

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

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

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

\(A=\frac{\sqrt{6+\sqrt{12}-\sqrt{8}-\sqrt{24}}}{\sqrt{2}+\sqrt{3}+1}\)

\(A=\frac{\sqrt{1+2+3+2\sqrt{3}-2\sqrt{2}-2\sqrt{2.3}}}{\sqrt{2}+\sqrt{3}+1}\)

\(A=\frac{\sqrt{\left(1-\sqrt{2}+\sqrt{3}\right)^2}}{\sqrt{2}+\sqrt{3}+1}=\frac{1-\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}+1}\)

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

\(A=\frac{\sqrt{3}-\sqrt{2}-1-\sqrt{6}+2+\sqrt{2}+3-\sqrt{6}-\sqrt{3}}{3-\left(\sqrt{2}+1\right)^2}\)

\(A=\frac{4-2\sqrt{6}}{3-3-2\sqrt{2}}=\frac{4-2\sqrt{3}}{-2\sqrt{2}}=\frac{2\left(2-\sqrt{3}\right)}{-2\sqrt{2}}=\frac{\sqrt{3}-2}{\sqrt{2}}=\frac{\sqrt{6}-2\sqrt{2}}{2}\)