b,
`(x-2)(x^2+2x+7)+2(x^2-4)-5(x-2)=0`
`⇔(x-2)(x^2+2x+7)+2(x-2)(x+2)-5(x-2)=0`
`⇔(x-2)[x^2+2x+7+2(x+2)-5]=0`
`⇔(x-2)(x^2+2x+7+2x+4-5)=0`
`⇔(x-2)(x^2+4x+6)=0`
`⇔(x-2)=0` (vì `x^2+4x+6>0∀x`)
`⇔x=2`
Vậy: `S={2}`
d,
`x^3+27+(x+3)(x-9)=0`
`⇔(x+3)(x^2-3x+9)+(x+3)(x-9)=0`
`⇔(x+3)(x^2-3x+9+x-9)=0`
`⇔(x+3)(x^2-2x)=0`
`⇔x(x+3)(x-2)=0`
`⇔`\(\left[ \begin{array}{l}x=0\\x+3=0\\x-2=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=-3\\x=2\end{array} \right.\)
Vậy: `S={0;-3;2}`
e, `8x^3-50x=0`
`⇔2x(4x^2-25)=0`
`⇔2x(2x-5)(2x+5)=0`
`⇔`\(\left[ \begin{array}{l}2x=0\\2x-5=0\\2x+5=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\2x=5\\2x=-5\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{array} \right.\)
Vậy: `S={0;\frac{5}{2};-\frac{5}{2}}`
f,
`2(x+3)-x^2-3x=0`
`⇔2(x+3)-(x^2+3x)=0`
`⇔2(x+3)-x(x+3)=0`
`⇔(x+3)(2-x)=0`
`⇔`\(\left[ \begin{array}{l}x+3=0\\2-x=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-3\\x=2\end{array} \right.\)
Vậy: `S={-3;2}`
