\(A=\dfrac{x}{x^2-4}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}=\dfrac{2}{\left(x-2\right)\left(x+2\right)}+\dfrac{6}{3\left(2-x\right)}+\dfrac{1}{x+2}\)
\(=\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{6\left(x+2\right)}{3\left(x-2\right)\left(x+2\right)}+\dfrac{3\left(x-2\right)}{3\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{3x-6x-12+3x-6}{3\left(x-2\right)\left(x+2\right)}=\dfrac{-18}{3\left(x-2\right)\left(x+2\right)}=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\)
b.
Ta có: \(A=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}=3\Leftrightarrow3x^2-12=-6\Leftrightarrow3x^2=6\Leftrightarrow x^2=2\Leftrightarrow x=\pm\sqrt{2}\) c. Hơi có vấn đề
c, bạn giải bất phương trình \(\dfrac{x+5}{2}+\dfrac{3-2x}{4}>x\)ra ta được \(x< \dfrac{13}{4}\)hay 3.25mà x là số nguyên lớn nhất \(\Rightarrow\)x=3
thay x=3 vào A ta đượcA=\(-\dfrac{6}{5}\)