Question

1) Cho biểu thức

B = ( \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\)) - \(\dfrac{x+2003}{x}\)

a) Tìm ĐKXĐ

b) Rút gọn B

c) Tìm giá trị nguyên của x để B có giá trị nguyên

Answer 1

\(a,ĐKXĐ;x\ne\pm1\)

\(a,\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right)-\dfrac{x+2003}{x}\)

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

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

\(=\dfrac{2.2x+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2003}{x}\)

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

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

\(=1-\dfrac{x+2003}{x}\)

\(=\dfrac{x-\left(x+2003\right)}{x}\)

\(=\dfrac{x-x-2003}{x}\)

\(=-\dfrac{2003}{x}\)

Answer 2

c/ để B nguyên ⇔ \(-\dfrac{2003}{x}\in Z\)

\(\Leftrightarrow x\inƯ\left(2003\right)\)

\(\Leftrightarrow x=\left\{\pm1;\pm2003\right\}\)

kết vợp với đkxđ

=> \(x=\left\{\pm2003\right\}\)

vậy.......