Tìm x > 1 biết:
\(5x\left(x-2000\right)-x+2000=0\)
Tìm x biết:
\(a,3x^3-3x=0\)
\(b,x\left(x-2\right)+x-2=0\)
\(c,5x\left(x-2000\right)-x+2000=0\)
a, 3x 3 - 3x = 0
=> 3x ( x 2 - 1 ) = 0
=> \(\orbr{\begin{cases}3x=0\\x^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}\Rightarrow[}\begin{cases}x=0\\x=1\\x=-1\end{cases}}\)
b, x ( x - 2 ) + ( x - 2 ) = 0
=> ( x - 2 ) ( x + 1 ) = 0
=> \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
c, 5x ( x - 2000 ) - x + 2000 = 0
=> ( x - 2000 ) ( 5x - 1 ) = 0
=> \(\orbr{\begin{cases}x-2000=0\\5x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2000\\x=\frac{1}{5}\end{cases}}}\)
Tìm x biết :
a) \(5x\left(x-2000\right)-x+2000=0\)
b) \(x^3-13x=0\)
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 \(x=-13\)
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 = 1515
Vậy x = 1515; 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
Tìm x, biết:
a)\(5x\left(x-2000\right)-x+2000=0\)
b)\(x^3-13x=0\)
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 ) \(x^3-13x=0\)
\(\Leftrightarrow x\left(x^2-13\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x^2-13=0\Rightarrow\left[\begin{array}{nghiempt}x=\sqrt{13}\\x=-\sqrt{13}\end{array}\right.\end{array}\right.\)
Tìm x, biết :
a, \(5x\left(x-2000\right)-x+2000=0\)
b, \(x^3-13x=0\)
Tìm x biết:
\(a,3x^3-3x=0\)
\(b,x\left(x-2\right)+x-2=0\)
\(c,5x\left(x-2000\right)-x+2000=0\)
a) \(3x^3-3x=0\)
\(\Rightarrow3x\left(x^2-1\right)=0\)
\(\Rightarrow3x\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{0;\pm1\right\}\)
b) \(x\left(x-2\right)+x-2=0\)
\(\Rightarrow x\left(x-2\right)+\left(x-2\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{-1;2\right\}\)
c) \(5x\left(x-2000\right)-x+2000=0\)
\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Rightarrow\left(5x-1\right)\left(x-2000\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x-2000=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2000\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{5};2000\right\}\)
Tìm x
a) \(5x\left(x-2000\right)-x+2000=0\)
b) \(x^3-13x=0\)
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 = \(-\sqrt{13}\)
tìm x biết \(5x\left(x-2000\right)-x+2000=0\)
phân tích đa thức thành nhân tử
\(14x^2y-21xy^2+28x^2y^2\)
\(10x\left(x-y\right)-8y\left(y-x\right)\)
tìm x>1 biết :5x X (X -2000) - X +2000=0
=> 5.x(x - 2000) - (x - 2000) = 0
=> (x - 2000) (5x - 1) = 0
=> x - 2000 = 0 => x = 2000
hoặc 5x - 1 = 0 => 5x = 1 => x = 1/5
Vậy x = 2000; x = 1/5
tìm x>1 biết:5x(x-2000)-x+2000=0
5x(x-2000)-x+2000=0
<=>5x(x-2000)-(x-2000)=0
<=>(5x-1)(x-2000)=0
<=>5x-1=0 hoặc x-2000=0
<=>5x=1 hoặc x=2000
5x=1,Mà x>1 =>loại
=>x=2000
Ta có:5\(\times\)(x-2000)-x+2000=0
x\(\times\)5-2000\(\times\)5-x+2000=0
x\(\times\)4-8000=0
\(\Rightarrow\)x\(\times\)4=8000
x=8000\(\div\)4=2000
Vậy x bằng 2000.