Đáp án:
Giải thích các bước giải:
a) `(x-1)(x^2+x+1)-x(x^2+1)=4`
`⇔ x^3+x^2+x-x^2-x-1-x^3-x=4`
`⇔ -1-x=4`
`⇔ x=-5`
Vậy `S={-5}`
b) `15x^2-5x=0`
`⇔ 5x(3x-1)=0`
`⇔` \(\left[ \begin{array}{l}5x=0\\3x-1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=\dfrac{1}{3}\end{array} \right.\)
Vậy `S={0;1/3}`
c) `(x-2)^2-3(x-2)=0`
`⇔ (x-2)(x-2)-3(x-2)=0`
`⇔ (x-2)(x-2-3)=0`
`⇔ (x-2)(x-5)=0`
`⇔` \(\left[ \begin{array}{l}x-2=0\\x-5=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
Vậy `S={2;5}`
d) `x^3-9x^2-x+9=0`
`⇔ x^2(x-9)-(x-9)=0`
`⇔ (x^2-1)(x-9)=0`
`⇔` \(\left[ \begin{array}{l}x^2-1=0\\x-9=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=1\\x=-1\\x=9\end{array} \right.\)
Vậy `S={1;-1;9}`
