Question

D = \(\frac{x-3}{x+1}-\frac{x+2}{x-1}+\frac{8x}{x^2-1}\)

a, Tìm điều kiện tiêu chuẩn và rút gọn bt D

b, Tìm x ∈ Z để D ∈ Z

c, Tìm x để D > 0 ; D < 0

Answer 1

ĐKXĐ: \(x\ne\pm1\)

\(D=\frac{\left(x-3\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{8x}{\left(x+1\right)\left(x-1\right)}\)

\(D=\frac{x^2-4x+3-\left(x^2+3x+2\right)+8x}{\left(x+1\right)\left(x-1\right)}=\frac{x+1}{\left(x+1\right)\left(x-1\right)}=\frac{1}{x-1}\)

Để \(D\in Z\Rightarrow1⋮\left(x-1\right)\Rightarrow x-1=Ư\left(1\right)=\left\{-1;1\right\}\)

\(\Rightarrow x=\left\{0;2\right\}\)

Để \(D>0\Rightarrow\frac{1}{x-1}>0\Rightarrow x-1>0\Rightarrow x>1\)

Để \(D< 0\Rightarrow x< 1\)