Đáp án+Giải thích các bước giải:
`c)`
`(x^2-1)(x+2)(x-3)=(x-1)(x^2-4)(x+5)`
`⇔(x^2-1)(x+2)(x-3)-(x-1)(x^2-4)(x+5)=0`
`⇔(x-1)(x+1)(x+2)(x-3)-(x-1)(x-2)(x+2)(x+5)=0`
`⇔[(x-1)(x+2)].[(x+1)(x-3)-(x-2)(x+5)]=0`
`⇔(x-1)(x+2).[x^2-2x-3-(x^2+3x-10)]=0`
`⇔(x-1)(x+2)(x^2-2x-3-x^2-3x+10)=0`
`⇔(x-1)(x+2)(7-5x)=0`
\(⇔\left[ \begin{array}{l}x-1=0\\x+2=0\\7-5x=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=1\\x=-2\\x=\dfrac{7}{5}\end{array} \right.\)
Vậy `x∈\{1;-2;7/5\}`
`d)`
`x^4+x^3+x+1=0`
`⇔x^3(x+1)+(x+1)=0`
`⇔(x+1)(x^3+1)=0`
`⇔(x+1)(x+1)(x^2-x+1)=0`
`⇔(x+1)^2(x^2-x+1)=0`
TH 1:
`(x+1)^2=0`
`⇔x+1=0`
`⇔x=-1`
TH 2:
`x^2-x+1=0`
`⇔x^2-2.x.(1)/2+1/4+3/4=0`
`⇔(x-1/2)^2+3/4=0`
`⇔(x-1/2)^2=-3/4` (Vô lí)
Vậy `x=-1`
