\(a,a=\sqrt{\dfrac{3}{5}}+\sqrt{\dfrac{5}{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{5}}+\dfrac{\sqrt{5}}{\sqrt{3}}\\ =\dfrac{\sqrt{3}.\sqrt{3}+\sqrt{5}.\sqrt{5}}{\sqrt{3}.\sqrt{5}}\\ =\dfrac{3+5}{\sqrt{15}}\\ =\dfrac{8}{\sqrt{15}}\)
Ta có:
\(15a^2=15.\left(\dfrac{8}{\sqrt{15}}\right)^2=15.\dfrac{64}{15}=64\)
\(8a\sqrt{15}=8.\dfrac{8}{\sqrt{15}}.\sqrt{15}=64\)
`=> 15a^2 -8a sqrt15 =64-64=0`
Ta có:
\(\sqrt{15a^2-8a\sqrt{15}+16}=\sqrt{0+16}=\sqrt{16}=4\)
\(b,\sqrt{a^2+2\sqrt{a^2-1}}-\sqrt{a^2-2\sqrt{a^2-1}}\\ =\sqrt{a^2-1+2\sqrt{a^2-1}+1}-\sqrt{a^2-1-2\sqrt{a^2-1}+1}\\ =\sqrt{\left(\sqrt{a^2-1}+1\right)^2}-\sqrt{\left(\sqrt{a^2-1}-1\right)^2}\\ =\sqrt{a^2-1}+1-\left|\sqrt{a^2-1}-1\right|\)
Thay `a=5` ta có:
\(\sqrt{5^2-1}+1-\left|\sqrt{5^2-1}-1\right|\\ =\sqrt{25-1}+1-\left|\sqrt{25-1}-1\right|\\ =\sqrt{24}+1-\left|\sqrt{24}-1\right|\\ =\sqrt{24}+1-\sqrt{24}+1\\ =2\)
