Question

Tìm x. biết: \(x^3-8=\left(x-2\right)^3\)

Answer 1

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

Theo đề ra \(x^3-8=\left(x-2\right)^3\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)=\left(x-2\right)^3\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4-\left(x-2\right)^2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4-x^2-4x-4\right)=0\)

\(\Leftrightarrow-2x\left(x-2\right)=0\)

\(\Leftrightarrow2x=0\) hoặc \(x-2=0\)

\(\Leftrightarrow x=0\) hoặc \(x=2\)

Vậy phương trình có tập nghiệm là S(0;2)\(\)

Answer 2

x_³-8 =x³-6x²+12x-8

⇔6x²-12x=0

⇔x²-2x=0

⇔x(x-2)=0

⇔\(|^{x=0}_{x=2}\)

vậy x=0 hoặc x=2

Answer 3

\(x^3-8=\left(x-2\right)^3\)

\(\Leftrightarrow x^3-8-x^3+3x^2-3x+8=0\)

\(\Leftrightarrow3x\left(x-1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}3x=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)