giải phươnh trình;a,(x+1)(2x-3)=(x-2)mũ2 b,x mũ 2-4-5(x-2)mũ2=0 24/10/2021 Bởi Hadley giải phươnh trình;a,(x+1)(2x-3)=(x-2)mũ2 b,x mũ 2-4-5(x-2)mũ2=0
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 $ $→(x – 2)[x + 2 – 5(x – 2)]=0$ $→(x – 2)(x + 2 – 5x +10)=0$ $→ (x – 2)(- 4x + 12)=0$ $→\left[ \begin{array}{l}x-2=0\\-4x+12=0\end{array} \right.$ $→\left[ \begin{array}{l}x=2\\x=3\end{array} \right.$ $Vậy$ $S=${$2;3$} Bình luận
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 ↔ \(\left[ \begin{array}{l}x_{1}=\frac{-5+\sqrt[]{41}}{2}\\x_{2}=\frac{-5-\sqrt[]{41}}{2}\end{array} \right.\) Bình luận
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 $
$→(x – 2)[x + 2 – 5(x – 2)]=0$
$→(x – 2)(x + 2 – 5x +10)=0$
$→ (x – 2)(- 4x + 12)=0$
$→\left[ \begin{array}{l}x-2=0\\-4x+12=0\end{array} \right.$
$→\left[ \begin{array}{l}x=2\\x=3\end{array} \right.$
$Vậy$ $S=${$2;3$}
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
↔ \(\left[ \begin{array}{l}x_{1}=\frac{-5+\sqrt[]{41}}{2}\\x_{2}=\frac{-5-\sqrt[]{41}}{2}\end{array} \right.\)