Đáp án:
`f)S={1;3/4}`
`g)S={2;1}`
`h)S={2;3}`
`i)S={-2;1}`
Giải thích các bước giải:
`f)(x-1)(x²+5x-2)-x³+1=0`
`⇔x³+5x²-2x-x²-5x+2-x³+1=0`
`⇔(x³-x³)+(5x²-x²)-(2x+5x)+(2+1)=0`
`⇔4x²-7x+3=0`
`⇔4x²-4x-3x+3=0`
`⇔4x(x-1)-3(x-1)=0`
`⇔(x-1)(4x-3)=0`
`⇔`$\left[\begin{matrix} x-1=0\\ 4x-3=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=1\\ x=\dfrac{3}{4}\end{matrix}\right.$
Vậy `S={1;3/4}`
`g)x²-3x+2=0`
`⇔x²-2x-x+2=0`
`⇔x(x-2)-(x-2)=0`
`⇔(x-2)(x-1)=0`
`⇔`$\left[\begin{matrix} x-2=0\\ x-1=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=2\\ x=1\end{matrix}\right.$
Vậy `S={2;1}`
`h)x³-8x²+21x-18=0`
`⇔x³-2x²-6x²+12x+9x-18=0`
`⇔(x³-2x²)-(6x²-12x)+(9x-18)=0`
`⇔x²(x-2)-6x(x-2)+9(x-2)=0`
`⇔(x-2)(x²-6x+9)=0`
`⇔(x-2)(x-3)²=0`
`⇔`$\left[\begin{matrix} x-2=0\\ (x-3)²=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=2\\ x-3=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=2\\ x=3\end{matrix}\right.$
Vậy `S={2;3}`
`i)x^4+x²+6x-8=0`
`⇔x^4+x³-x³+4x²-x²-2x²+4x+2x-8=0`
`⇔(x^4-x³+4x²)+(x³-x²+4x)-(2x²-2x+8)=0`
`⇔x²(x²-x+4)+x(x²-x+4)-2(x²-x+4)=0`
`⇔(x²+x-2)(x²-x+4)=0`
`⇔(x²+2x-x-2)(x²-x+1/4+15/4)=0`
`⇔[x(x+2)-(x+2)][(x²-x+1/4)+15/4]=0`
`⇔(x+2)(x-1)[(x-1/2)^2+15/4]=0`
Ta có:`(x-1/2)^2≥0∀x`
`⇒(x-1/2)^2+15/4≥15/4>0∀x`
`⇒` vô nghiệm
`⇔(x+2)(x-1)=0`
`⇔`$\left[\begin{matrix} x+2=0\\ x-1=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=-2\\ x=1\end{matrix}\right.$
Vậy `S={-2;1}`
