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 = $\frac{1}{5}$

Vậy x = $\frac{1}{5}$; 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 =

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 ) \(5x\left(x-2000\right)-x+2000=0\)

\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)

\(\Leftrightarrow\left(x-2000\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}5x-1=0\\x-2000=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{5}\\x=2000\end{array}\right.\)

b ) \(x^3-13x=0\)

\(\Leftrightarrow x\left(x^2-13\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x^2-13=0\Rightarrow\left[\begin{array}{nghiempt}x=\sqrt{13}\\x=-\sqrt{13}\end{array}\right.\end{array}\right.\)

b) 5x(x-2000)-x+2000=0

\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\\ \Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+2000\\5x=0+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\5x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)

Ai giúp minh làm bài 5 phía trên với

c) Ta có: \(2x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-3}{2}\end{matrix}\right.\)

d) Ta có: \(5x^2+x=0\)

\(\Leftrightarrow x\left(5x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{5}\end{matrix}\right.\)

a,\(5x\left(x-2000\right)-x+2000=0\)

\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)

\(\Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)

b,\(x^3-13x=0\)

\(\Rightarrow x.x^2-13x=0\)

\(\Rightarrow x\left(x^2-13\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2=13\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\end{matrix}\right.\)

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=\(\sqrt{13}\)

a) \(5x\left(x-2000\right)-x+2000=0\)

\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)

\(\Leftrightarrow\left(x-2000\right)\left(5x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2000=0\\5x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2000\\x=\frac{1}{5}\end{cases}}\)

b) \(x^3-13x=0\)

\(\Leftrightarrow x\left(x^2-13\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=13\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{cases}}\)

do không viết được dấu ngoặc vuông nên mình thay bằng dấu ngoặc nhọn nha!

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;\(\sqrt{13}\)

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

=> \(\orbr{\begin{cases}x-2000=0\\5x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2000\\x=\frac{1}{5}\end{cases}}}\)

a ) \(5x\left(x-2000\right)-x+2000=0\)

\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)

\(\Leftrightarrow\left(x-2000\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)

Vậy \(x=2000\) và \(x=\dfrac{1}{5}\)

b ) \(x^3-13x=0\)

\(\Leftrightarrow x\left(x^2-13\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\sqrt{13}\end{matrix}\right.\)

Vậy \(x=0\) và \(x=\sqrt{13}\)

c ) \(x+5x^2=0\)

\(\Leftrightarrow x\left(1+5x\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\1+5x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\)

Vậy \(x=0\) và \(x=-\dfrac{1}{5}\)

d ) \(\left(x+1\right)=\left(x+1\right)^2\)

\(\Leftrightarrow\left(x+1\right)-\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)\left[1-\left(x+1\right)\right]=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy \(x=0\) và \(x=-1\)

e ) \(x^3+x=0\)

\(\Leftrightarrow x\left(x^2+1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x^2+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\\left(loại\right)\end{matrix}\right.\)

Vậy \(x=0\)

a, \(5x\left(x-2000\right)-x+2000=0\)

\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x-2000\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\x-2000=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2000\end{matrix}\right.\)

b,\(x^3-13x=0\)

\(\Leftrightarrow x\left(x ^2-13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\end{matrix}\right.\)

c,\(x+5x^2=0\)

\(\Leftrightarrow x\left(5x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\)

d,\(x+1=\left(x+1\right)^2\)

\(\Leftrightarrow\left(x+1\right)-\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)\left(1-x-1\right)=0\)

\(\Leftrightarrow-x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

e,\(x^3+x=0\)

\(\Leftrightarrow x\left(x^2+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

CHÚC BẠN HỌC TỐT........

Ở câu e, cho mình sửa lại:

\(x^2+1=0\) (vô lý, do \(x^2+1\ge0\))

Vậy x=0