\(113^2-26.113+13^2\)

\(=\left(113-13\right)^2\)

\(=100^2\)

\(=10000\)

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

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

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

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 

.....