Question

(x+2)(x²-2x+4)-(x³+2x²)=0 

(x-1³)+(2-x)(4+2x+x²)+3x(x+2)=17

giúp mik nha :>

Answer 1

\(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+2x^2\right)=0\\ \Rightarrow x^3+8-x^3-2x^2=0\\ \Rightarrow-2x^2+8=0\Rightarrow x^2=4\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ \left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\\ \Rightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\\ \Rightarrow9x=10\\ \Rightarrow x=\dfrac{10}{9}\)

Answer 2

\(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+2x^2\right)=0\)

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

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

\(4-x^2=0\)

\(x^2=4\)

⇒\(\left[{}\begin{matrix}x^2=\left(-2\right)^2\\x^2=2^2\end{matrix}\right.\)

⇒\(\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)