Giải thích các bước giải:
$A=(4x^5+4x^4-5x^3+5x-2)^{2014}+2015$
$=[4x^4(x+1)-5x(x^2-1)-2]^{2014}+2015$
$=[4x^4(x+1)-5x(x+1)(x-1)-2]^{2014}+2015$
$=[(x+1)(4x^4-5x^2+5x)-2]^{2014}+2015$
$=[x(x+1)(4x^3-5x+5)-2]^{2014}+2015$
$x=\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}$
$=\dfrac{1}{2}\dfrac{\sqrt{(\sqrt{2}-1)^2}}{2-1}$
$=\dfrac{1}{2}.\dfrac{\sqrt{2}-1}{1}$
$=\dfrac{\sqrt{2}-1}{2}$
$\text{Thay $x=\dfrac{\sqrt{2}-1}{2}$ vào $A$, ta có:}$
$\{\dfrac{\sqrt{2}-1}{2}.(\dfrac{\sqrt{2}-1}{2}+1)\dfrac{\sqrt{2}-1}{2}.[4(\dfrac{\sqrt{2}-1}{2})^3-5\dfrac{\sqrt{2}-1}{2}+5]-2\}^{2014}+2015$
$=\{\dfrac{1+\sqrt{2}}{2}.\dfrac{\sqrt{2}-1}{2}.[\dfrac{-7+5\sqrt{2}}{2}-\dfrac{5}{2}(\sqrt{2}-1)+5]-2\}^{2014}+2015$
$=(\dfrac{1}{4}.4-2)^{2014}+2015$
$=(-1)^{2014}+2015$
$=2016$
Học tốt!!!
