Question

[︵✰Ah]

(2017-x)\(^{\text{3}}\) + (2019-x)\(^{\text{3}}\) + (2x-4036)\(^{\text{3}}\) =0 

Tìm x

Answer 1

[Nguyễn Lê Phước Thịnh]

Đặt 2017-x=a; 2019-x=b

\(\Leftrightarrow a+b=4036-2x\)

\(\Leftrightarrow-\left(a+b\right)=2x-4036\)

Phương trình trở thành: \(a^3+b^3-\left(a+b\right)^3=0\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)-\left(a+b\right)^3=0\)

\(\Leftrightarrow-3ab\left(a+b\right)=0\)

mà -3<0

nên \(ab\left(a+b\right)=0\)

\(\Leftrightarrow\left(2017-x\right)\left(2019-x\right)\left(4036-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2017-x=0\\2019-x=0\\4036-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=2019\\x=2018\end{matrix}\right.\)

Vậy: S={2017;2018;2019}

Answer 2

[Trần Mạnh]

Cho \(\left(2017-x\right)^3=x;\left(2019-x\right)^3=y;\left(2x-4036\right)^3=z\)

Ta có: \(x+y+z=0\)

\(=>x+y=-z\) \(=>\left(x+y\right)^3=-z^3\)

Ta có: \(x^3+y^3+z^3=\left(x+y\right)^3-3xy\left(x+y\right)+z^3=-z^3-3xy\left(-z\right)+z^3=3xyz\)

Vì (2017-x)³ + (2019-x)³ + (2x-4036)³ =0 

=>\(3\left(2017-x\right)\left(2019-x\right)\left(2x-4036\right)=0\)

Gải phương trình được x=2017; x=2019; x=2018