\(2011\equiv1\left(mod2010\right)\Rightarrow2011^{2009}\equiv1\left(mod2010\right)\)

\(2009\equiv-1\left(mod2010\right)\Rightarrow2009^{2011}\equiv-1\left(mod2010\right)\)

\(\Rightarrow2009^{2011}+2011^{2009}\equiv0\left(mod2010\right)\Rightarrow2009^{2011}+2011^{2009}⋮2010\)

mod là sao

ai giai nhanh mk se hanng ngay cho

(x+1)(x+4)(x+2)(x+3)-8

= (x^2 +5x +4)(x^2 + 5x+6)-8

Đặt x^2 +5x +5= a

Ta có : (x^2 +5x+4)(x^2 +5x+6)-8

=(a-1)(a+1)-8

= a^2 -1 -8

=a^2 -9 

= a^2 - 3^2

= (a-3)(a+3)

=(x^2 +5x+2)(x^2 + 5x+7)

ĐKXĐ: \(x\ne1\)

\(A=\frac{5x+1}{x^3-1}-\frac{1-2x}{x^2+x+1}-\frac{2}{1-x}\)

\(A=\frac{5x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{\left(1-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{5x+1-x+1+2x^2-2x+2x^2+2x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{4x^2+4x+4}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{4\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{4}{x-1}\left(x^2+x+1\ne0\right)\)

b, ĐỂ x nhân giá trị nguyên 

\(\Rightarrow4⋮x-1\)

\(\Rightarrow x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Nếu : x - 1 = 1 => x = 2 

  x - 1 = -1 => x = 0 

x - 1 = 2 => x = 3 

.....

\(4x\left(x-2007\right)-\left(x-2007\right)=0\)

\(=\left(4x-1\right)\left(x-2007\right)=0\)

\(\orbr{\begin{cases}4x-1=0\Rightarrow4x=1\Rightarrow x=\frac{1}{4}\\x-2007=0\Rightarrow x=2007\end{cases}}\)

kl : x = 1/4 hoặc 2007

\(4x\left(x-2007\right)-x+2007=0\)

\(4x\left(x-2007\right)-\left(x-2007\right)=0\)

\(\left(x-2007\right)\left(4x-1\right)=0\)

\(\leftrightarrow\orbr{\begin{cases}x-2007=0\\4x-1=0\end{cases}}\leftrightarrow\orbr{\begin{cases}x=2007\\x=\frac{1}{4}\end{cases}}\)

vậy \(x\in\left\{2007;\frac{1}{4}\right\}\)

ta có : \(2x^2-x+1⋮2x+1\)

\(\Rightarrow\left(2x^2+x\right)-\left(2x-1\right)+2⋮2x+1\)

\(\Rightarrow x\left(2x+1\right)-\left(2x+1\right)+2⋮2x+1\)

\(\left(2x+1\right)\left(x-1\right)+2⋮2x+1\)

\(\Rightarrow2⋮2x+1\Rightarrow2x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Nếu : 2x + 1 = 1 => x = 0  ( TM ) 

    2x + 1 = -1 => x = -1  ( TM ) 

  2x + 1 = 2 => x = 3/2 ( loại ) 

2x + 1 = -2 => x = -3/2  ( loại ) 

\(\Rightarrow x\in\left\{0;-1\right\}\)

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

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

\(=x\left(6x+5-3y\right)\)

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

\(=\left(6x^2-3xy\right)+\left(10x-5y\right)\)

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

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

\(x^3+2x^2+3x=0\)\(\Leftrightarrow x.\frac{x^3+2x^2+3x}{x}=0\)

\(\Leftrightarrow x\left(x^2+2x+3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+2x+3=0\end{cases}}\)

Ta sẽ c/m \(x^2+2x+3=0\) vô nghiệm.Thật vậy:

\(x^2+2x+3=\left(x+1\right)^2+2\ge2\forall x\)

Từ đó suy ra \(x^2+2x+3=0\) vô nghiệm.

Vậy : x = 0

\(\left(x+2\right)\left(2x-1\right)+1=4x^2\)

\(2x^2-x+4x-2+1=4x^2\)

\(\Rightarrow2x^2-3x+1=0\)

\(2x\left(x-1\right)-\left(x-1\right)=0\)

\(\left(x-1\right)\left(2x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)

ý còn lại tham khảo bài tth