Đáp án:
a) `x=-2/7`
`b)x=0` hoặc `x=-26/3`
`c)x=0` hoặc `x=±1`
Giải thích các bước giải:
`a) (x+2)(x^2-2x+4)-x(x^2+2)=15`
`⇔x^3+8-x^3-2x-15=0`
`⇔-2x-7=0`
`⇔x=-2/7`
Vậy `x=-2/7`
`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=0`
`⇔3x^2+26x=0`
`⇔x(3x+26)=0`
`⇔`\(\left[ \begin{array}{l}x=0\\3x+26=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=-\dfrac{26}{3}\end{array} \right.\)
Vậy `x=0` hoặc `x=-26/3`
`c) (x^2-1)^3-(x^4+x^2+1)(x^2-1)=0`
`⇔x^6-3x^4+3x^2-1-x^6+1=0`
`⇔3x^2-3x^4=0`
`⇔3x^2(1-x^2)=0`
`⇔3x^2(1-x)(1+x)=0`
`⇔`\(\left[ \begin{array}{l}3x^2=0\\1-x=0\\x+1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=±1\end{array} \right.\)
Vậy `x=0` hoặc `x=±1`
