Bài 1:
\(a,=\dfrac{x^2-5x+6+x-3+x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\\ =\dfrac{x^2-3x+2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\\ =\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x-3}\\ b,=\dfrac{3x+9+2x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{2x}\\ =\dfrac{\left(x^2+2x+9\right)\left(x+3\right)}{2x\left(x-3\right)}\)
Bài 2:
\(a,ĐK:x\ne\pm1\\ b,A=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\\ A=\dfrac{10\cdot4}{2\cdot5}=4\left(đpcm\right)\)
Bài 3:
\(B=\dfrac{x\left(x^3+3x^2+3x+1\right)}{2\left(x^2+2x+1\right)}=\dfrac{x\left(x+1\right)^3}{2\left(x+1\right)^2}=\dfrac{x\left(x+1\right)}{2}\)
Ta thấy \(x\left(x+1\right)\) là tích 2 số liên tiếp nên sẽ chẵn với mọi \(x\ne-1\)
\(\Leftrightarrow x\left(x+1\right)⋮2\)
Vậy \(B\in Z,\forall x\ne-1\)
