(x+4y)3=-2(5x-y)

3x+12y=-10x+2y

13x=-10y

\(x+\left(x+1\right)+\left(x+2\right)+....+\left(x+2003\right)=2004\)

\(\Leftrightarrow x+\left(x+x+....+x\right)+\left(1+2+....+2003\right)=2004\)

\(\Leftrightarrow x+2003x+2007006=2004\)

\(\Leftrightarrow2004x=2004-2007006\)

\(\Leftrightarrow2004x=-2005002\)

\(\Leftrightarrow x=-\frac{2001}{2}=-1000,5\)

(x² - 2)(x² + 2x + 2)

\(\Leftrightarrow2\left(x+1\right)^3=56\Leftrightarrow\left(x+1\right)^3=28\Leftrightarrow\)

Ta có: \(\dfrac{\left(x+3\right)\left(x-3\right)}{3}+2=x\left(1-x\right)\)

\(\Leftrightarrow\dfrac{x^2-9}{3}+\dfrac{6}{3}=\dfrac{3x\left(1-x\right)}{3}\)

\(\Leftrightarrow x^2-9+6=3x-3x^2\)

\(\Leftrightarrow x^2-3-3x+3x^2=0\)

\(\Leftrightarrow4x^2-3x-3=0\)

\(\Delta=9-4\cdot4\cdot\left(-3\right)=9+48=57\)

Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là 

\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{57}}{8}\\x_2=\dfrac{3+\sqrt{57}}{8}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3-\sqrt{57}}{8};\dfrac{3+\sqrt{57}}{8}\right\}\)

\(\left(x+5\right)\left(2x-5\right)=\left(x+3\right)\left(2x-1\right)\)

\(2x^2-5x+10x-25=2x^2-x+6x-3\)

\(2x^2+5x-25=2x^2+5x-3\)

\(2x^2+5x-25-2x^2-5x+3=0\)

\(-25+3=0\)

\(-22=0\)  (Vô lí)

⇒ không tồn tại x

Ta có: (x+1).(3-x)=0

=> x+1=0 hoặc 3-x =0

+) x+1=0            +) 3-x=0

   x= 0-1                    x=3-0

    x=-1                       x=3

Vậy x = -1 hoặc x=3

( x + 1 )( 3 - x ) = 0

=> x + 1 = 0 hoặc 3 - x = 0

=> x = -1 hoặc x = 3

Vậy x = -1 hoặc x = 3

`(x+1).(3-x)=0`

`=>`\(\left[\begin{array}{l}x+1=0\\3-x=0\end{array}\right.\)

`=>\(\left[\begin{array}{l}x=0-1\\x=3-0\end{array}\right.\)

`=>`\(\left[\begin{array}{l}x=-1\\x=3\end{array}\right.\)