Giải nhanh giúp tớ với nha , cảm ơn rấc nhiều

Đáp án: $A=20172018$

Giải thích các bước giải:

Ta có:

$A=a^3+a^2c-abc+b^2c+b^3+20172018$

$\to A=(a^3+b^3)+(a^2c+b^2c)-abc+20172018$

$\to A=(a^3+b^3)+c(a^2+b^2)-abc+20172018$

$\to A=(a+b)^3-3ab(a+b)+c((a+b)^2-2ab)-abc+20172018$

$\to A=(a+b)^3-3ab(a+b)+c(a+b)^2-2abc-abc+20172018$

$\to A=(a+b)^3-3ab(a+b)+c(a+b)^2-3abc+20172018$

$\to A=(a+b)^3+c(a+b)^2-3ab(a+b)-3abc+20172018$

$\to A=(a+b)^2(a+b+c)-3ab(a+b+c)+20172018$

$\to A=((a+b)^2-3ab)(a+b+c)+20172018$

Ta có:

$a+b+c=\sqrt{2015-\sqrt{4029}}-\sqrt8+\sqrt{18}-\sqrt{2015+\sqrt{4029}}$

$\to a+b+c=\sqrt{2015-\sqrt{4029}}-2\sqrt2+3\sqrt{2}-\sqrt{2015+\sqrt{4029}}$

$\to a+b+c=\sqrt{2015-\sqrt{4029}}+\sqrt{2}-\sqrt{2015+\sqrt{4029}}$

$\to a+b+c=\sqrt{2015-\sqrt{4029}}-\sqrt{2015+\sqrt{4029}}+\sqrt{2}$

$\to a+b+c=\dfrac1{\sqrt2}\cdot (\sqrt{4030-2\sqrt{4029}}-\sqrt{4030+2\sqrt{4029}})+\sqrt{2}$

$\to a+b+c=\dfrac1{\sqrt2}\cdot (\sqrt{(\sqrt{4029}-1)^2}-\sqrt{(\sqrt{4029}+1)^2})+\sqrt{2}$

 $\to a+b+c=\dfrac1{\sqrt2}\cdot ((\sqrt{4029}-1)-(\sqrt{4029}+1))+\sqrt{2}$

 $\to a+b+c=\dfrac1{\sqrt2}\cdot (-2)+\sqrt{2}$

$\to a+b+c=-\sqrt2+\sqrt2$

$\to a+b+c=0$

$\to A=((a+b)^2-3ab)\cdot 0+20172018$

$\to A=20172018$