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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: a) 2(x-3) ²= 7x ²- 63 b) 3x ³ +2x ² +2x -3 =0

Tháng Hai 11, 2023 Toán học 2 Comments 0 views

Toán Lớp 8: a) 2(x-3) ²= 7x ²- 63
b) 3x ³ +2x ² +2x -3 =0

Comments ( 2 )

  1. thanhthuythu
    thanhthuythu
    11 Tháng Hai, 2023 at 10:00 chiều
    Reply
    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}

  2. thudan
    thudan
    11 Tháng Hai, 2023 at 10:00 chiều
    Reply
    Giải đáp:
    $\\$
    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}
    $\\$
    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}

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222-9+11+12:2*14+14 = ? ( )

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