Question

Tìm x biết:

a) \(\left(x-3\right)^2\) - x(\(x^2\) - 1) + \(9x^2\) = 4

b) \(\left(x+2\right)^3\) - \(\left(x-2\right)^3\) - 4x(3x - 1) = -8

c) \(\left(x-2018\right)^3\) + \(\left(x-2020\right)^3\) + \(\left(4038-2x\right)^3\) = 0

Answer 1

a: Sửa đề: \(\left(x-3\right)^3-x\left(x^2-1\right)+9x^2=4\)

=>\(x^3-9x^2+27x-27-x^3+x+9x^2=4\)

=>28x=31

=>\(x=\dfrac{31}{28}\)

b: \(\left(x+2\right)^3-\left(x-2\right)^3-4x\left(3x-1\right)=-8\)

=>\(x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)-12x^2+4x=-8\)

=>\(x^3-6x^2+16x+8-x^3+6x^2-12x+8=-8\)

=>4x+16=-8

=>4x=-24

=>x=-6

c: Đặt x-2018=a; x-2020=b

=>a+b=x-2018+x-2020=2x-4038

\(\left(x-2018\right)^3+\left(x-2020\right)^3+\left(4038-2x\right)^3=0\)

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

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

=>-3ab(a+b)=0

=>ab(a+b)=0

=>(x-2018)(x-2020)(4038-2x)=0

=>\(\left[{}\begin{matrix}x-2018=0\\x-2020=0\\4038-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2018\\x=2020\\x=2019\end{matrix}\right.\)