Vì a là nghiệm của phương trình : \(4x^2+x\sqrt{2}-\sqrt{2}=0\)
nên : \(4a^2+\sqrt{2}x-\sqrt{2}=0\)
\(\Rightarrow2\sqrt{2}a^2+a-1=0\)
\(\Rightarrow\)\(a^2=\dfrac{1-a}{2\sqrt{2}}\)
Ta có :\(\sqrt{a^4+a+1}=\sqrt{\dfrac{\left(1-a\right)^2}{8}+a+1}=\sqrt{\dfrac{1-2a+a^2+8a+8}{8}}=\sqrt{\dfrac{a^2+6a+9}{8}}=\sqrt{\dfrac{\left(a+3\right)^2}{8}}=\dfrac{a+3}{2\sqrt{2}}\)(do a>0)
Thay \(\sqrt{a^4+a+1}=\dfrac{a+3}{2\sqrt{2}}\) vào biểu thức cần tính, ta được :
\(\dfrac{a+1}{\sqrt{a^4+a+1}-a^2}=\dfrac{\left(a+1\right)\left(\sqrt{a^4+a+1}+a^2\right)}{\left(\sqrt{a^4+a+1}-a^2\right)\left(\sqrt{a^4+a+1}-a^2\right)}=\dfrac{\left(a+1\right)\left(\sqrt{a^4+a+1}\right)}{a^4+a+1-a^4}=\dfrac{\left(a+1\right)\left(\dfrac{a+3}{2\sqrt{2}}+a^2\right)}{a+1}=\dfrac{a+3+2\sqrt{2}a^2}{2\sqrt{2}}=\dfrac{\left(2\sqrt{2}a^2+a-1\right)+4}{2\sqrt{2}}=\dfrac{4}{2\sqrt{2}}=\sqrt{2}\)
