Toán mn giúp e vs ạ a) x^2 – 3x + 2 = x – 2 b) x(x-5) = 4 – 2(3x + 1) 06/10/2021 By Gabriella mn giúp e vs ạ a) x^2 – 3x + 2 = x – 2 b) x(x-5) = 4 – 2(3x + 1)
a) x^2 – 3x + 2 = x – 2 <=> x^2 – 3x + 2 – x + 2 = 0 <=> x^2 – 4x + 4 = 0 <=> ( x – 2 )^2 = 0 <=> x = 2 b) x(x-5) = 4 – 2(3x + 1) <=> x^2-5x-4+6x+2=0 <=> x^2+x-2=0 <=> x^2-x+2x-2=0 <=> x(x-1)+2(x-1)=0 <=> (x-1)(x+2)=0 <=> x-1=0 hoặc x+2=0 <=> x=1 hoặc x=-2 Trả lời
Đáp án: Giải thích các bước giải: $a) x² – 3x + 2 = x – 2$ $⇔ x² – 3x + 2 – x – 2 = 0 $ $⇔x² – 4x=0$ $⇔x(x-4)=0$ ⇔ \(\left[ \begin{array}{l}x=0\\x-4=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=0\\x=4\end{array} \right.\) $b) x(x-5) = 4 – 2(3x + 1)$ $⇔ x² – 5x = 4 – 6x – 2$ $⇔x² – 5x – 4 + 6 x + 2=0$ $⇔ x² + x – 2+0$ Ta có: $a+b+c = 1+1+(-2)=0$ ⇔\(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\) Trả lời
a) x^2 – 3x + 2 = x – 2
<=> x^2 – 3x + 2 – x + 2 = 0
<=> x^2 – 4x + 4 = 0
<=> ( x – 2 )^2 = 0
<=> x = 2
b) x(x-5) = 4 – 2(3x + 1)
<=> x^2-5x-4+6x+2=0
<=> x^2+x-2=0
<=> x^2-x+2x-2=0
<=> x(x-1)+2(x-1)=0
<=> (x-1)(x+2)=0
<=> x-1=0 hoặc x+2=0
<=> x=1 hoặc x=-2
Đáp án:
Giải thích các bước giải:
$a) x² – 3x + 2 = x – 2$
$⇔ x² – 3x + 2 – x – 2 = 0 $
$⇔x² – 4x=0$
$⇔x(x-4)=0$
⇔ \(\left[ \begin{array}{l}x=0\\x-4=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\x=4\end{array} \right.\)
$b) x(x-5) = 4 – 2(3x + 1)$
$⇔ x² – 5x = 4 – 6x – 2$
$⇔x² – 5x – 4 + 6 x + 2=0$
$⇔ x² + x – 2+0$
Ta có: $a+b+c = 1+1+(-2)=0$
⇔\(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\)