Toán Lớp 8: Tìm x biết: a) x2 + 5x = 0 b) 3x(x – 1) = 1 – x
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Toán Lớp 8: Tìm x biết: a) x2 + 5x = 0 b) 3x(x – 1) = 1 – x
Comments ( 2 )
Giải đáp + Lời giải và giải thích chi tiết:
a)
x^2 +5x=0
⇔x(x+5)=0
⇔ $\left[\begin{matrix} x=0\\ x+5=0\end{matrix}\right.$
⇔ $\left[\begin{matrix} x=0\\ x=-5\end{matrix}\right.$
Vậy x=0 hoặc x=-5
b)
3x(x-1)=1-x
⇔ 3x(x-1)-1+x=0
⇔ 3x(x-1)+(x-1)=0
⇔ (x-1)(3x+1)=0
⇔ $\left[\begin{matrix} x-1=0\\ 3x+1=0\end{matrix}\right.$
⇔ $\left[\begin{matrix} x=1\\ x=\dfrac{-1}{3}\end{matrix}\right.$
Vậy x=1 hoặc x=-1/3
$a)$ x^2 + 5 x = 0
⇒x (x + 5) = 0
⇒(x + 5/2)^2 – 25/4 = 0
\(⇒\left[ \begin{array}{l}x=\dfrac{-5+5}{2\cdot \:1}\\x=\dfrac{-5-5}{2\cdot \:1}\end{array} \right.\)
\(⇒\left[ \begin{array}{l}x=0\\x=-5\end{array} \right.\)
Vậy $x=0,\:x=-5$
$b)$ $3x\left(x-1\right)=1-x$
$⇒3x^2-3x=1-x$
$⇒3x^2-3x+x=1-x+x$
$⇒3x^2-2x=1$
$⇒3x^2-2x-1=1-1$
$⇒3x^2-2x-1=0$
\(⇒\left[ \begin{array}{l}x=\dfrac{-\left(-2\right)+4}{2\cdot \:3}\\x=\dfrac{-\left(-2\right)-4}{2\cdot \:3}\end{array} \right.\)
\(⇒\left[ \begin{array}{l}x=1\\x=-\dfrac{1}{3}\end{array} \right.\)
Vậy $x=1,\:x=-\dfrac{1}{3}$