Question

  em đang cần gấp ạa huhu

Answer 1

a: ĐKXĐ: \(x\notin\left\{-3;2\right\}\)

b: \(B=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)

\(=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\)

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

\(=\dfrac{x^2-4-x-8}{\left(x-2\right)\left(x+3\right)}=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)

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

c: \(B=\dfrac{-3}{4}\)

=>\(\dfrac{x-4}{x-2}=\dfrac{-3}{4}\)

=>4(x-4)=-3(x-2)

=>4x-16=-3x+6

=>7x=22

=>\(x=\dfrac{22}{7}\left(nhận\right)\)

d: \(x^2-9=0\)

=>\(x^2=9\)

=>\(\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Thay x=3 vào B, ta được:

\(B=\dfrac{3-4}{3-2}=\dfrac{-1}{1}=-1\)

e: Để B là số nguyên thì \(x-4⋮x-2\)

=>\(x-2-2⋮x-2\)

=>\(-2⋮x-2\)

=>\(x-2\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{3;1;4;0\right\}\)