Rút gọn A
\(A=\dfrac{\left(x-1\right)^2}{x^2-4x+3}\)
\(=\dfrac{\left(x-1\right)^2}{x^2-x-3x+3}\)
\(=\dfrac{\left(x-1\right)^2}{x\left(x-1\right)-3\left(x-1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x-3\right)}\)
\(=\dfrac{x-1}{x-3}\)
Vậy \(A=\dfrac{x-1}{x-3}\)
Để \(A< 1\) thì
\(\dfrac{x-1}{x-3}< 1\)
\(\Leftrightarrow\dfrac{x-1-x+3}{x-3}< 0\)
\(\Leftrightarrow2< 0\left(VL\right)\)
Vậy \(S=\varnothing\)
-ĐKXĐ: \(x\ne1;x\ne3\)
\(A=\dfrac{\left(x-1\right)^2}{x^2-4x+3}< 1\)
\(\Leftrightarrow\dfrac{\left(x-1\right)^2}{x^2-x-3x+3}< 1\)
\(\Leftrightarrow\dfrac{\left(x-1\right)^2}{x\left(x-1\right)-3\left(x-1\right)}< 1\)
\(\Leftrightarrow\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x-3\right)}< 1\)
\(\Leftrightarrow\dfrac{x-1}{x-3}< 1\)
\(\Leftrightarrow\dfrac{x-1}{x-3}-\dfrac{x-3}{x-3}< 0\)
\(\Leftrightarrow\dfrac{x-1-x+3}{x-3}< 0\)
\(\Leftrightarrow\dfrac{2}{x-3}< 0\)
\(\Leftrightarrow x-3< 0\)
\(\Leftrightarrow x< 3\)
-Vậy tập nghiệm của BĐT là {x l x<3}
