\(x.\left(x-2009\right)-2010x+2009.2010=0\)

\(x.\left(x-2009\right)-2010\left(x-2009\right)=0\)

\(\left(x-2009\right)\left(x-2010\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2009=0\\x-2010=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2009\\x=2010\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=2009\\x=2010\end{cases}}\)

Đề sai rồi bạn. Đây là phương trình vô tỉ mà?

\(x\left(x-2009\right)-2010x+2009\times2010=0\)

\(x^2-2009x-2010x+2009\times2010=0\)

\(x\left(x-2010\right)-2009\left(x-2010\right)=0\)

\(\left(x-2009\right)\left(x-2010\right)=0\)

nên x - 2009 = 0 

x = 2009

x-2010=0

x=2010

b: 

1: \(\Leftrightarrow2x\left(x+2\right)=0\)

=>x=0 hoặc x=-2

11x² - 4x = 0

=> x(11x - 4) = 0

Có 2 TH xảy ra :

TH1 : x = 0

TH2 : 11x - 4 = 0 => 11x = 4 => x = 4/11

=>x(11x-4)=0
=> x=0
     11x-4=0
=> x=0
     x=\(\frac{4}{11}\)

12.x²-4x=0

<=>12.x².x=0+4

=>12.x³=4

x³=4:12

x³=1/3

=>x.x.x=1/3

=>x=1/3 hoặc x=1

1)3x⁴-5x³y+6x²-10xy+2

=(3x⁴-5x³y)+(6x²-10xy)+2

=x³(3x-5y)+2x(3x-5y)+2

=x³.0+2x.0+2

=0+0+2

=2

2) x⁵-2010x⁴+2010x³-2010x²+2010x-2020

=x⁵-(2009+1)x⁴+(2009+1)x³-(2009+1)x²+(2009+1)x-2009-11

=x⁵-(x+1)x⁴+(x+1)x³-(x+1)x²+(x+1)x-x-11

=x⁵-x⁵-x⁴+x⁴+x³-x³-x²+x²+x-x-11

=-11

2, Với x= 2009 => 2010=x+1

=> \(x^5-2010\text{x}^4+2010\text{x}^3-2010\text{x}^2+2010\text{x}-2020=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-2020\)

\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-2020\)

\(=x-2020\)

\(=2009-2020\\ =-11\)

#) TL :

a) x² - 5x + 6 = 0                                              

   x² - 2x - 3x + 6 = 0                                              

  x( x - 2) - 3( x - 2 ) = 0

 ( x - 3)( x -2 ) = 0

=> \(\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\)

=>\(\orbr{\begin{cases}x=3\\x=2\end{cases}}\)

b) Đag bí :)

Chúc bn hok tốt :3

b) \(x^2+11x+10=0\)

\(\Leftrightarrow x^2+10x+x+10=0\)

\(\Leftrightarrow x\left(x+10\right)+\left(x+10\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+10\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-10\end{cases}}\)

a) \(x^2-5x+6=0\)

\(\left(a=1;b=-5;c=6\right)\)

\(\Delta=b^2-4ac=\left(-5\right)^2-4.1.6=1>0\)

\(x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-\left(-5\right)+\sqrt{1}}{2.1}=3\)

\(x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-\left(-5\right)-\sqrt{1}}{2.1}=2\)

Vậy x1 = 3 ; x2 = 2

b) \(x^2+11x+10=0\)

\(\left(a=1;b=11;c=10\right)\)

\(\Delta=b^2-4ac=11^2-4.1.10=81>0\)

\(\sqrt{\Delta}=\sqrt{81}=9\)

\(x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-11+9}{2.1}=-1\)

\(x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-11-9}{2.1}=-10\)

Vậy x1 = -1 ; x2 = -10 

HỌC TỐT !!!