Đáp án:
Giải thích các bước giải:
Bài 6:
`-8x^3+12x^2-6x=63`
`⇔ -8x^3+12x^2-6x-63=0`
`⇔ -(8x^3-12x^2+6x+63)=0`
`⇔ -(8x^3-24x^2+12x^2+42x-36x+63)=0`
`⇔ -(2x+3)(4x^2-12x+21)=0`
Ta có: `4x^2-12x+21= 4x^2-12x+9+12=(2x-3)^2+12 \ge 12∀x`
`⇔ 2x+3=0`
`⇔ x=-3/2`
Vậy `S={-3/2}`
Bài 7:
`x^3-4x^2+4x=0`
`⇔ x(x^2-4x+4)=0`
`⇔ x(x-2)^2=0`
`⇔` \(\left[ \begin{array}{l}x=0\\(x-2)^2=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\)
Vậy `S={0;2}`
Bài 8:
`25x^3-10x^2+x=0`
`⇔ x(25x^2-10x+1)=0`
`⇔ x(5x-1)^2=0`
`⇔` \(\left[ \begin{array}{l}x=0\\(5x-1)^2=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=\dfrac{1}{5}\end{array} \right.\)
Vậy `S={0;1/5}`
