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b) 5x(x-2000)-x+2000=0
\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\\ \Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+2000\\5x=0+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\5x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
5x(x – 2000) – x + 2000 = 0
⇔ 5x(x – 2000) – (x – 2000) = 0
(Có x – 2000 là nhân tử chung)
⇔ (x – 2000).(5x – 1) = 0
⇔ x – 2000 = 0 hoặc 5x – 1 = 0
+ x – 2000 = 0 ⇔ x = 2000
+ 5x – 1 = 0 ⇔ 5x = 1 ⇔ x = 1/5.
Vậy có hai giá trị của x thỏa mãn là x = 2000 và x = 1/5.
![](https://rs.olm.vn/images/avt/0.png?1311)
5x.(x-2000)-x+2000=0
=> 5x.(x-2000)-(x-2000)=0
=> (x-2000)-(5x-1)=0
=> x-2000=0 => x=2000
Hoặc
=> 5x-1=0 => 5x=1 => x=1:5 => x=1/5
Vậy x=2000 hoặc x=1/5.
\(5x.\left(x-2000\right)-x+2000=0\)
\(\Rightarrow5x.\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Rightarrow\left(x-2000\right).\left(5x-1\right)=0\)
\(\orbr{\begin{cases}x-2000=0\\5x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2000\\x=\frac{1}{5}\end{cases}}\)
Vậy x=2000 hoặc x=\(\frac{1}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 5x(x - 2000) - x + 2000 = 0
=> 5x(x - 2000) - (x - 2000) = 0
=> (x - 2000).(5x - 1) = 0
=> x - 2000 = 0 hoặc 5x - 1 = 0
=> x = 2000 hoặc 5x = 1
=> x = 2000 hoặc x = 1/5
b) x3 - 13x = 0
=> x.(x2 - 13) = 0
=> x = 0 hoặc x2 - 13 = 0
=> x = 0 hoặc x2 = 13, vô lí
=> x = 0
a) 5x(x-2000)-(x-2000)=(5x-1)(x-2000)=0 nên x=1/5 hoặc x=2000
b)\(x^3-13x=x\left(x^2-13\right)=0\)\(\Rightarrow\)x=0 hoặc x^2=13 hay x=\(\sqrt{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 5x(x - 2000) - (x - 2000) = 0
tương đương (x - 2000)(5x - 1) = 0
tương đương x = 2000 hoặc x = 1/5
b) x(x^2 -13) = 0
\(x\left(x-\sqrt{13}\right)\left(x+\sqrt{13}\right)=0\)
tương đương x = 0 hoặ x = \(\sqrt{13}\)hoặc x = \(-\sqrt{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài giải:
a) 5x(x -2000) - x + 2000 = 0
5x(x -2000) - (x - 2000) = 0
(x - 2000)(5x - 1) = 0
Hoặc 5x - 1 = 0 => 5x = 1 => x =
Vậy x = ; x = 2000
b) x3 – 13x = 0
x(x2 - 13) = 0
Hoặc x = 0
Hoặc x2 - 13 = 0 => x2 = 13 => x = ±√13
Vậy x = 0; x = ±√13
a) 5x(x-2000)-x+2000=0
5x(x-2000)-(x-2000)=0
(x-2000)(5x-1)=0
\(\Leftrightarrow\) x-2000=0 hoặc 5x-1=0
\(\Leftrightarrow\) x=2000 hoặc x=\(\dfrac{1}{5}\)
b) \(x^3-13x=0\)
\(x\left(x^2-13\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(x^2-13=0\)
\(\Leftrightarrow x=0\) hoặc \(x=13\) hoặc \(x=-13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 5x(x-2000)-x+2000=0
<=>5x(x-2000)-(x-2000)=0
<=>(5x-1)(x-2000)=0
<=>5x-1=0 hoặc x-2000=0
<=>x=1/5 hoặc x=2000
b, x3-13x=0
<=>x(x2-13)=0
<=>x=0 hoặc x2-13=0
<=>x=0 hoặc x=\(\sqrt{13}\) hoặc x=\(-\sqrt{13}\)
a,5x(x-2000)-x+2000=0
=>5x(x-2000)-(x-2000)=0
=>(5x-1)(x-2000)=0
=>x-2000=0 hoặc 5x-1=0
=>x=2000 hoặc x=1/5
vậy x=1/5;2000
b,x3-13x=0
=>(x2-13)x=0
=>x2-13=0 hoặc x=0
=>x=0 hoặc x=\(\sqrt{13}\)
vậy x=0;\(\sqrt{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) \(5x\left(x-2000\right)-x+2000=0\)
\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Leftrightarrow\left(x-2000\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}5x-1=0\\x-2000=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{5}\\x=2000\end{array}\right.\)
b ) \(x^3-13x=0\)
\(\Leftrightarrow x\left(x^2-13\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x^2-13=0\Rightarrow\left[\begin{array}{nghiempt}x=\sqrt{13}\\x=-\sqrt{13}\end{array}\right.\end{array}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Rightarrow\left(x-2000\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\\ b,\Rightarrow x\left(x^2-13\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\\ c,\Rightarrow3x\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ d,\Rightarrow\left(x-5\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\\ e,\Rightarrow\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Cậu chỉ cần bỏ ngoặc đi thôi rùi làm tiếp
5x2 - 10000x - x +2000 = 0
\(\Leftrightarrow5x^2-10001x+2000=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2000\\x=0,2\end{matrix}\right.\)