a) \(A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x^2-3x+2}\)
\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x^2-x-2x+2}\)
\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x\left(x-1\right)-2\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{\left(x-1\right)\left(x-2\right)}\)
\(\Leftrightarrow A=\frac{\left(4x-1\right)\left(x-1\right)-\left(x-3\right)\left(x-2\right)-2x+4}{\left(x-2\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{4x^2-4x-x+1-x^2+2x+3x-6-2x+4}{\left(x-2\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{3x^2-2x-1}{\left(x-2\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{3x^2-3x+\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}\)\(=\frac{3x\left(x-1\right)+\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}\)\(=\frac{\left(x-1\right)\left(3x+1\right)}{\left(x-2\right)\left(x-1\right)}\)\(=\frac{3x+1}{x-2}\)
b)\(\frac{3x+1}{x-2}=\frac{3x-6+7}{x-2}=\frac{3x-6}{x-2}+\frac{7}{x-2}=3+\frac{7}{x-2}\)
Ta có : \(x-2\inƯ_7\left\{-7;-1;1;7\right\}\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-2=-7\\x-2=-1\\x-2=1\\x-2=7\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}\text{x=-5}\\\text{x=1}\\\text{x=3}\\\text{x}=9\end{array}\right.\)
\(\text{x}=1\) (loại)
Vậy giá trị nguyên tập hợp x là:
x=-5;3;9
