tìm x biết a)x/2+x^2/8=0 b)4-x=2(x-4)^2 c)(x^2+1)(2x-1)+2x=4

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1. a, $\frac{x}{2}$ + $\frac{x²}{8}$=0

   <=> $\frac{4x}{8}$ + $\frac{x²}{8}$=0

   <=> 4x+ x²=0

   <=> x(x+4)=0

   TH1: x=0

   TH2: x+4=0 <=> x=-4

   Vậy x ∈ {0;-4}

   b, 4-x=2(x-4)²

   <=> 2(x-4)²+ (x-4)=0

   <=> (x-4)(2x-8+1)=0

   <=> (x-4)(2x-7)=0

   TH1: x-4=0 <=> x=4

   TH2: 2x-7=0 <=> x=7/2

   Vậy x∈ {4;7/2}

   c, (x²+1)(2x-1)+2x=4

   <=> 2x³- x²+ 2x-1 +2x-4=0

   <=> 2x³- x²+ 4x-5=0

   <=> 2x³- 2x²+ x²- x+ 5x-5=0

   <=> 2x²(x-1)+ x(x-1)+ 5(x-1)=0

   <=> (x-1)(2x²+x+5)=0

   <=> x-1= 0 (vì 2x²+x+5= 2(x+1/4)²+ 39/8 luôn lớn hơn 0)

    <=> x=1

   Vậy x=1

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2. a)x/2+x^2/8=0

   ⇔ 4x/8 + x²/8 = 0

   ⇔ $\frac{x^2+4x}{8}$ = 0

   ⇔ x² + 4x = 0

   ⇔ x(x+4)=0

   ⇔ \(\left[ \begin{array}{l}x=0\\x+4=0\end{array} \right.\) 

   ⇔ \(\left[ \begin{array}{l}x=0\\x=-4\end{array} \right.\) 

   b)4-x=2(x-4)^2

   ⇔ 4 -x = 2(x² – 8x + 16)

   ⇔ 4 – x = 2x² – 16x + 32

   ⇔ 4 – 32 = 2x² – 16x + x

   ⇔ -28 = 2x² – 15x

   ⇒ 2x² – 15x + 28 = 0

   ⇔ (x-4)(2x-7)=0

   ⇔ \(\left[ \begin{array}{l}x-4=0\\2x-7=0\end{array} \right.\) 

   ⇔ \(\left[ \begin{array}{l}x=4\\x=7/2\end{array} \right.\) 

   c)(x^2+1)(2x-1)+2x=4

   ⇔ 2x³ – x² + 2x – 1 + 2x = 4

   ⇔ 2x³ – x² + 2x – 1 + 2x – 4 = 0

   ⇔ 2x³ – x² + 4x – 5 = 0

   ⇔ (x-1)(2x² + x + 5) = 0

   Vì 2x² + x + 5 ≥ 0 ⇒ x – 1 = 0 ⇒ x  =1

    

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