a) Khi x = 3 thì : \(K=\frac{2.3+7}{3+1}=\frac{6+7}{4}=\frac{13}{4}\)

b)\(K=\frac{2x+7}{x+1}=\frac{2x+2+5}{x+1}=\frac{2\left(x+1\right)+5}{x+1}=2+\frac{5}{x+1}\)

Để K là số nguyên thì : \(5⋮x+1\Leftrightarrow x+1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\Leftrightarrow x\in\left\{-6;-2;0;4\right\}\)

c) \(K=\frac{2x+7}{x+1}=1\Leftrightarrow2x+7=x+1\Leftrightarrow x+6=0\Leftrightarrow x=-6.\)

a) Với x = -3

=> K = \(\frac{2.\left(-3\right)+7}{-3+1}=\frac{-6+7}{-2}=-\frac{1}{2}\)

b) Ta có:

K = \(\frac{2x+7}{x+1}=\frac{2\left(x+1\right)+5}{x+1}=2+\frac{5}{x+1}\)

Để K \(\in\)Z  <=> \(5⋮x+1\) <=> \(x+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Lập bảng :

Vậy ...

c)Ta có: K = 1

=> \(\frac{2x+7}{x+1}=1\)

=> \(2x+7=x+1\)

=> \(2x-x=1-7\)

=> \(x=-6\)

                                                         Bài giải

                           Ta có : \(K=\frac{2x+7}{x+1}\)

a, Thay x = - 3 ta được : 

\(K=\frac{2x+7}{x+1}=\frac{2\cdot\left(-3\right)+7}{-3+1}=\frac{-6+7}{-2}=\frac{-1}{-2}=\frac{1}{2}\)

b, Ta có : 

               \(K=\frac{2x+7}{x+1}=\frac{x+1+x+1+5}{x+1}=\frac{\left(x+1\right)+\left(x+1\right)+5}{x+1}=\frac{2\left(x+1\right)}{x+1}+\frac{5}{x+1}=2+\frac{5}{x+1}\)

\( \text{Để K là số nguyên thì }5\text{ }⋮\text{ }x+1\)

                                                       \(\Leftrightarrow\text{ }x +1\inƯ\left(5\right)=\left\{\pm1\text{ ; }\pm5\right\}\)

Ta có bảng :

\(\text{Vậy }K\in Z\text{ khi }x\in\left\{-2\text{ ; }0\text{ ; }-6\text{ ; }4\right\}\)

c, Nếu K = 1 thì ta có :

\(K=\frac{2x+7}{x+1}=1\)

\(\Rightarrow\text{ }2x+7=x+1\)

       \(2x-x=-7+1\)

 \(\Rightarrow\text{ }x=-6\)

em ko biết vì em mới học lớp 5 

k cho em nha

a) để B là phân số

=> 2x-1\(\ne\)0

=>2x\(\ne\)1

=>x\(\ne\)\(\frac{1}{2}\)

b) sửa đề :Tìm x để B có giá trị là  1 số nguyên

để B nguyên => x\(\in\)Z

=> 2x+5\(⋮\)2x-1

ta có : 2x-1\(⋮\)2x-1

=>(2x-5)-(2x-1)\(⋮\)2x-1

=>-4\(⋮\)2x-1

=>2x-1\(\in\)Ư(-4)={\(\pm1;\pm2;\pm4\)}

ta có bảng :

Mà x \(\in Z\)

nên x\(\in\){1;0}

a)ĐKXĐ:

\(x-1\ne0;x+1\ne0;x\ne0\)

\(\Leftrightarrow x\ne1;x\ne-1;x\ne0\)

b)\(K=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right).\frac{x+2003}{x}\)

\(=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{x+2003}{x}\)

\(=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}+\frac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{x+2003}{x}\)

\(=\frac{x^2+2x+1+x^2-2x+1+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)

\(=\frac{3x^2-4x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)

\(=\frac{3x^2-3x-x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)

\(=\frac{3x.\left(x-1\right)-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)

\(=\frac{\left(x-1\right)\left(3x-1\right)}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)

\(=\frac{\left(3x-1\right)\left(x+2003\right)}{\left(x+1\right).x}\)

\(=\frac{3x^2+6008x-2003}{x^2+x}\)

câu c bí

1.3+(-2)+x=5

-1+x=5

x=5-(-1)

x=6

Nhớ tick cho mình nha

Ta có: \(f\left(x\right)=x^2+px+q\)

\(\Rightarrow f\left(f\left(x\right)+x\right)=\left(f\left(x\right)+x\right)^2+p\left(f\left(x\right)+x\right)+q\)

\(=f\left(x\right)^2+2f\left(x\right).x+x^2+p.f\left(x\right)+p.x+q\)

\(=f\left(x\right)^2+2f\left(x\right).x+p.f\left(x\right)+\left(x^2+p.x+q\right)\)

\(=f\left(x\right)^2+2f\left(x\right).x+p.f\left(x\right)+f\left(x\right)\)

\(=f\left(x\right).\left(f\left(x\right)+2x+p+1\right)=f\left(x\right).\left(x^2+px+q+2x+p+1\right)\)

\(=f\left(x\right).\left(\left(x+1\right)^2+\left(x+1\right)p+q\right)=f\left(x\right).f\left(x+1\right)\)

Vậy tồn tại số nguyên k để f(k) = f(2008).f(2009) ( Chọn x = 2018 thì \(k=f\left(2018\right)+2018\))