Question

Cho biểu thức : A=\(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

a) Rút gọn A

b) Tìm x để A >0

c) Tìm x ∈ Z để A nguyên dương

Answer 1

a) ĐKXĐ: x∉{-3;2}

Ta có: \(A=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)

Answer 2

a,\(A=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(x-2\right)}\)-\(\frac{5}{\left(x-2\right)\left(x+3\right)}\)-\(\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)dkxdx-2≠0,x+3≠0,x≠2,-3

\(=\frac{x^2-4-5-\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)

=\(\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)

=\(\frac{\left(x+3\right)\left(x-4\right)}{\left(x-2\right)\left(x+3\right)}\)

=\(\frac{x-4}{x-2}\)