`a)2(x-3)²=7x²-63`

`⇔2(x²-6x+9)-7x²+63=0`

`⇔2x²-12x+18-7x²+63=0`

`⇔(2x²-7x²)+12x+(18+63)=0`

`⇔-5x²+12x+81=0`

`⇔-5(x²-12/5x-81/5)=0`

`⇔-5(x²-3x+27/5x-81/5)=0`

`⇔-5[x(x-3)+27/5(x-3)]=0`

`⇔-5(x-3)(x+27/5)=0`

`⇔(x-3)(x+27/5)=0`

`⇔`\(\left[ \begin{array}{l}x-3=0\\x+\dfrac{27}{5}=0\end{array} \right.\)

`⇔`\(\left[ \begin{array}{l}x=3\\x=-\dfrac{27}{5}\end{array} \right.\)

Vậy `S={3;-27/5}`

`b)3x³+2x²+2x+3=0`

`⇔(3x³+3)+(2x²+2x)=0`

`⇔3(x³+1)+2x(x+1)=0`

`⇔3(x+1)(x²-x+1)+2x(x+1)=0`

`⇔(x+1)[3(x²-x+1)+2x]=0`

`⇔(x+1)(3x²-3x+3+2x)=0`

`⇔(x+1)(3x²-x+3)=0`

`⇔`\(\left[ \begin{array}{l}x+1=0\\3x²-x+3=0\end{array} \right.\)

`⇔`\(\left[ \begin{array}{l}x=-1\\3x²-x+3=0(vô nghiệm)\end{array} \right.\)

 Vậy `S={-1}`

Đáp án:

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`a,`

`2 (x-3)^2 = 7x^2 - 63`

`↔ 2 (x^2 - 6x + 9) - 7x^2 + 63=0`

`↔ 2x^2 - 12x + 18 - 7x^2 + 63=0`

`↔ (2x^2 - 7x^2) - 12x + (18 + 63)=0`

`↔ -5x^2 - 12x + 81 =0`

`↔ -5 [x^2 + 12/5x - 81/5] = 0`

`↔ -5 [ (x^2 - 3x) + (27/5x - 81/5)]=0`

`↔ -5 [x (x-3) + 27/5 (x-3)]=0`

`↔ -5 [(x-3) (x+27/5)]=0`

`↔ (x-3) (x+27/5)=0`

`↔` \(\left[ \begin{array}{l}x-3=0\\x+\dfrac{27}{5}=0\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}x=0+3\\x=0-\dfrac{27}{5}\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}x=3\\x=\dfrac{-27}{5}\end{array} \right.\) 

Vậy `S ∈ {3; (-27)/5}`

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`b,`

`3x^3 + 2x^2 + 2x + 3 = 0`

`↔ (3x^3 + 3) + (2x^2+2x)=0`

`↔ 3 (x^3 + 1) + 2x (x + 1) = 0`

`↔ 3 (x^3 + 1^3) + 2x (x+1)=0`

`↔ 3 (x +  1) (x^2 - x +1) + 2x (x+1)=0`

`↔ (x+1) [3 (x^2-x+1) + 2x]=0`

Trường hợp 1 :

`↔x+1=0`

`↔x=0-1`

`↔x=-1`

Trường hợp 2 :

`↔3 (x^2 - x+1) + 2x=0`

`↔ 3x^2 - 3x + 3 + 2x=0`

`↔ 3x^2 + (-3x+2x) + 3=0`

`↔3x^2 - x + 3 = 0`

`↔ 3 [x^2 - 1/3x + 1] = 0`

`↔ 3 [x^2 - 2 . 1/6x + 1/36 + 35/36] =0`

`↔ 3 [x^2 - 2 . 1/6 + (1/6)^2 + 35/36]=0`

`↔ 3 (x - 1/6)^2 + 35/12=0`

Với mọi `x` có : `(x-1/6)^2` $\geqslant 0$

`↔ 3 (x-1/6)^2` $\geqslant 0 ∀x$

`↔ 3 (x-1/6)^2 + 35/12` $\geqslant$ `35/12 \ne 0`

`->` Đa thức vô nghiệm

Vậy `S ∈ {-1}`