Ta có : `E=`$\dfrac{x^2+8}{x^2+2}$
`=` $\dfrac{x^2+2+6 }{x^2+2}$
`=` ${\dfrac{x^2+2}{x^2+2} + \dfrac{6}{x^2+2}}$
`=1+`$\dfrac{6}{x^2+2}$
Với mọi x ta luôn có : ${ x^{2} \geq 0 }$
`=> x^{2} + 2 \geq 2`
`=>` ${\dfrac{6}{x^2+2} \leq \dfrac{6}{2}=3}$
`=> 1 + ` ${\dfrac{6}{x^2+2} \leq 1+3}$
`=> 1+ ` ${\dfrac{6}{x^2+2} \leq 4 }$
Dấu `=` xảy ra `<=> x^2=0`
`<=> x = 0 `
Vậy `Max_E=4 <=> x =0`
