a/ (do (x-3)^2 + 1 ≠0 vs mọi x

a) (x-3)³-3+x=0

=> (x-3)³+(x-3)=0

=> (x-3)(x²-6x+10)

=> \(\left[{}\begin{matrix}x-3=0\\x^2-6x+10=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=3\\\left(x-3\right)^2=1\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=3\\x=4\\x=2\end{matrix}\right.\)

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: \(5x^2+x=0\)

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

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

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.

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=\(\frac{1}{5}\)

Ta có:

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

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

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

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

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

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

(x-2000)(5x-1) =0

x -2000 = 0=> x=2000

5x-1 =0 => x = 1/5

Ta có: 5x(x-2000)-x+2000=0

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

(x-2000). (5x-1)=0

TH1: x-2000=0

x=2000

TH2: 5x-1= 0

x=1/5

Vậy x=2000 hoặc x=1/5

=> 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

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

Supper Saydan Goku

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.