Register Now

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

## Toán Lớp 8: tìm x 5x + x^2 = 0 2(x+5) – x^2 – 5x = 0

Toán Lớp 8: tìm x
5x + x^2 = 0
2(x+5) – x^2 – 5x = 0

1. Giải đáp + giải thích các bước giải:
5x + x^2 = 0
<=> x(x + 5) = 0
<=> $$\left[ \begin{array}{l}x=0\\x+5=0\end{array} \right.$$
<=>$$\left[ \begin{array}{l}x=0\\x=-5\end{array} \right.$$
Vậy x \in {0 ; -5}
2(x + 5) – x^2 – 5x = 0
<=> 2(x + 5) – (x^2 + 5x) = 0
<=> 2(x + 5) – x(x + 5) = 0
<=> $$\left[ \begin{array}{l}2 – x = 0\\x+5=0\end{array} \right.$$
<=>$$\left[ \begin{array}{l}x= 2\\x=-5\end{array} \right.$$
Vậy x \in {2 ; -5}

2. 1)
$5x+x^2=0$
$⇔ x^2+5x=0$
$⇔x_{1,\:2}=\frac{-5\pm \sqrt{5^2-4.1.0}}{2.1}$
$⇔x_{1,\:2}=\frac{-5\pm \:5}{2.1}$
$⇔x_1=\frac{-5+5}{2.1},\:x_2=\frac{-5-5}{2.1}$
$$⇔\left[ \begin{array}{l}x=0\\x=-5\end{array} \right.$$
2)
$2\left(x+5\right)-x^2-5x=0$
$⇔ -x^2-3x+10=0$
$⇔x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\left(-1\right)\cdot \:10}}{2\left(-1\right)}$
$⇔x_{1,\:2}=\frac{-\left(-3\right)\pm \:7}{2\left(-1\right)}$
$⇔x_1=\frac{-\left(-3\right)+7}{2\left(-1\right)},\:x_2=\frac{-\left(-3\right)-7}{2\left(-1\right)}$
$$⇔\left[ \begin{array}{l}x=-5\\x=2\end{array} \right.$$