ta có x+y=\(\sqrt{10}\)=>(x+y)^2=10
A=(x^4+1)(y^4+1)
=x^4.y^4+1+x^4+y^4+2x^2.y^2-2x^2.y^2
=x^4.y^4+1+(x^2+y^2)^2-2x^y^2=x^4.y^4+1+[(x+y)^2-2xy]
=x^4.y^4+1+(10-2xy)-2x^2.y^2
=x^4.y^4+1+100-40xy+4.x^2.y^2-2x^2.y^2
=x^4.y^4+101-40xy+2.x^2.y^2
=(x^4.y^4-8.x^2.y^2+16)+(10.x^2.y^2-40xy+40)+45
=(x^2.y^2-4)^2+10.(xy-2)^2+45\(\ge\)0
dấu = xảy ra \(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y=\sqrt{10}\\x.y=2\end{matrix}\right.\)
vậy Min A=45
\(\left\{{}\begin{matrix}x+y=\sqrt{10}\\x.y=2\end{matrix}\right.\)là nghiệm pt x^2-\(\sqrt{10}\)x+2
=>\(\Delta\)=(-\(\sqrt{10}\))^2-4.2=2>0
=>\(\left\{{}\begin{matrix}x=\dfrac{\sqrt{10}-\sqrt{2}}{2}\\y=\dfrac{\sqrt{10}+\sqrt{2}}{2}\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=\dfrac{\sqrt{10}-\sqrt{2}}{2}\\y=\dfrac{\sqrt{10}+\sqrt{2}}{2}\end{matrix}\right.\)
ta có x+y=√1010=>(x+y)^2=10
A=(x^4+1)(y^4+1)
=x^4.y^4+1+x^4+y^4+2x^2.y^2-2x^2.y^2
=x^4.y^4+1+(x^2+y^2)^2-2x^y^2=x^4.y^4+1+[(x+y)^2-2xy]
=x^4.y^4+1+(10-2xy)-2x^2.y^2
=x^4.y^4+1+100-40xy+4.x^2.y^2-2x^2.y^2
=x^4.y^4+101-40xy+2.x^2.y^2
=(x^4.y^4-8.x^2.y^2+16)+(10.x^2.y^2-40xy+40)+45
=(x^2.y^2-4)^2+10.(xy-2)^2+45≥45
dấu = xảy ra ⇔⇔{x+y=√10x.y=2{x+y=10x.y=2
vậy Min A=45
