giải phươnh trình;a,(x+1)(2x-3)=(x-2)mũ2 b,x mũ 2-4-5(x-2)mũ2=0

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1. Đáp án:

   $a, S =$ {$\frac{√37 – 3}{2} ; \frac{-√37 – 3}{2}$}

   $b, S =${$2;3$}

   Giải thích các bước giải:

   $a, (x + 1)(2x – 3) = (x – 2)²$

   $→ 2x² – x – 3 = x² – 4x + 4$

   $→ 2x² – x – 3 – x² + 4x – 4 = 0$ 

   $→ x²+ 3x – 7 = 0 $

   $→ (x² + 3x + \frac{9}{4}) – \frac{37}{4} = 0$

   $→ (x² + 2.x.\frac{3}{2} + \frac{9}{4}) = \frac{37}{4}$

   $→ (x + \frac{3}{2})² = (± \frac{√37}{2})$

   $→\left[ \begin{array}{l}x+ \frac{3}{2} = \frac{√37}{2}\\x+ \frac{3}{2} = \frac{-√37}{2}\end{array} \right.$

   $→\left[ \begin{array}{l}x=\frac{√37-3}{2}\\x=\frac{-√37 – 3}{2}\end{array} \right.$

   $Vậy$ $S=${$\frac{√37 – 3}{2} ;\frac{-√37-3}{2}$}

   $b, x² – 4 – 5(x – 2)² = 0$

   $→ (x – 2)(x+2) – 5(x-2)² = 0 $

   $→(x – 2)[x + 2 – 5(x – 2)]=0$

   $→(x – 2)(x + 2 – 5x +10)=0$

   $→ (x – 2)(- 4x + 12)=0$

   $→\left[ \begin{array}{l}x-2=0\\-4x+12=0\end{array} \right.$

   $→\left[ \begin{array}{l}x=2\\x=3\end{array} \right.$

   $Vậy$ $S=${$2;3$}

    

   Trả lời

2. a, (x+1)(2x-3)=$(x-2)^{2}$

   ↔ $2x^{2}$ – 3x + 2x – 3 = $x^{2}$ – 4x + 4

   ↔ $x^{2}$ + 3x – 7 = 0

   ↔ \(\left[ \begin{array}{l}x_{1}=\frac{-3+\sqrt[]{37}}{2}\\x_{2}=\frac{-3-\sqrt[]{37}}{2}\end{array} \right.\)

   b, $x^{2}$ – 4 – 5$(x-2)^{2}$ = 0

   ↔ $x^{2}$ – 4 – 5($x^{2}$ – 4x + 4) =0

   ↔ $x^{2}$ – 4 – 5$x^{2}$ – 20x + 20 =0

   ↔ $-4x^{2}$ – 20x + 16 =0

   ↔ \(\left[ \begin{array}{l}x_{1}=\frac{-5+\sqrt[]{41}}{2}\\x_{2}=\frac{-5-\sqrt[]{41}}{2}\end{array} \right.\)

   Trả lời