Đáp án:
Giải thích các bước giải:
a) `(2x-1)^2-25=0`
`⇔ (2x-1)^2-(5)^2=0`
`⇔ (2x-1-5)(2x-1+5)=0`
`⇔ (2x-6)(2x+4)=0`
`⇔` \(\left[ \begin{array}{l}2x-6=0\\2x+4=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=3\\x=-2\end{array} \right.\)
Vậy `S={3;-2}`
b) `8x^3-50x=0`
`⇔ 2x(4x^2-25)=0`
`⇔2x(2x-5)(2x+5)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\x=-\dfrac{5}{2}\\x=\dfrac{5}{2}\end{array} \right.\)
Vậy `S={0;5/2;-5/2}`
c) `(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+2x+4-5)=0`
`⇔ (x-2)(x^2+4x+6)=0`
Do `x^2+4x+6=(x+2)^2+2 \ge 2 \forall x`
`⇔ x-2=0`
`⇔ x=2`
Vậy `S={2}`
