a) (x + 2) (x ² – 2x + 4) – x(x ²+ 2)=15
b) (x + 3) ³ – x (3x +1) ² + (2x +1) (4x ² – 2x +1)
c) (x ² -1) ³ – (x^4 + x ² + 1) (x ² – 1) =0
tìm x ạ
a) (x + 2) (x ² – 2x + 4) – x(x ²+ 2)=15
b) (x + 3) ³ – x (3x +1) ² + (2x +1) (4x ² – 2x +1)
c) (x ² -1) ³ – (x^4 + x ² + 1) (x ² – 1) =0
tìm x ạ
`a,(x+2)(x^2-2x+4)-x(x^2+2)=15`
`⇔x^3+8-x^3-2x=15`
`⇔-2x=7`
`⇔x=-7/2`
`b,(x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=28`
`⇔x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28`
`⇔3x^2+26x=0`
`⇔x(3x+26)=0`
`⇔x=0` hoặc `x=-26/3`
`c,(x^2-1)^3-(x^4+x^2+1)(x^2-1)=0`
`⇔(x^2-1)[(x^2-1)^2-x^4-x^2-1]=0`
`⇔(x-1)(x+1)(x^4-2x^2+1-x^4-x^2-1)=0`
`⇔-3x^2(x-1)(x+1)=0`
`⇔x=0` hoặc `x=+-1`
Vậy ………………….
`a,(x+2)(x^2-2x+4)-x(x^2+2)=15`
`⇔x^3+8-x^3-2x=15`
`⇔-2x=7`
`⇔x=-7/2`
`b,(x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=28`
`⇔x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28`
`⇔3x^2+26x=0`
`⇔x(3x+26)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\3x+26=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=0\\x=\dfrac{-26}{3}\end{array} \right.\)
`c,(x^2-1)^3-(x^4+x^2+1)(x^2-1)=0`
`⇔(x^2-1)[(x^2-1)^2-x^4-x^2-1]=0`
`⇔(x-1)(x+1)(x^4-2x^2+1-x^4-x^2-1)=0`
`⇔-3x^2(x-1)(x+1)=0`
`⇔` \(\left[ \begin{array}{l}x-1=0\\x+1=0\\-3x^2=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=±1\\x=0\end{array} \right.\)