Question

[Anh Nguyễn]

Các bạn giúp mình bài toán sau

\(\left(x+2\right)^3\text{-}\left(x+1\right)\left(x^2\text{-}x+1\right)=10\)

\(\left(x\text{-}1\right)^3\text{-}\left(x\text{-}2\right)\left(x^2+x+4\right).3x\left(x+1\right)=0\)

\(\left(x\text{-}3\right)^2\text{-}\left(x\text{-}2\right)\left(x^2+2x+4\right)\text{-}9x\left(x\text{-}1\right)=0\)

Answer 1

[✉lãɴH⃣╰❥häńZ͜͡╰❥k̫ınżζ●]

đề bài là tìm x à bạn? đề có cho điều kiện ko vậy ạ? (ví dụ như x nguyên?)

Answer 2

[✉lãɴH⃣╰❥häńZ͜͡╰❥k̫ınżζ●]

\(\left(x-1\right)^3+\left(x^3-8\right).3x.\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right).\left[\left(x-1\right)^2+\left(x^3-8\right).3x\right]=0\)

TH1: \(x-1=0\Leftrightarrow x=1\)

TH2: \(\left(x-1\right)^2+\left(x^3-8\right).3x=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x^3-8\right).3x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left\{{}\begin{matrix}x^3-8=0\\3x=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\\left\{{}\begin{matrix}x=2\\x=0\end{matrix}\right.\end{matrix}\right.\)

Vậy \(x\in\left\{0;1;2\right\}\)