1. Chứng tỏ cặp phân thức sau = nhau:
\(\dfrac{x^2-10x+21}{^{ }x^3-7x^2+x-7}\) và \(\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}\)
2. Điền vào chỗ trống những đa thức thích hợp:
a) \(\dfrac{....}{x^2+3x+2}\)=\(\dfrac{3x^2+4x-4}{3x^2+7x+2}\) b) \(\dfrac{5x^2+7x-2}{2x+1}=\dfrac{5x^3-8x^2-23x+6}{...}\)
c) \(\dfrac{...}{6x^2+8x+2}=\dfrac{3x^2+4x-4}{3x^2+7x+2}\)
Câu 1:
\(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{\left(x-7\right)\left(x-3\right)}{\left(x-7\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)
\(\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}=\dfrac{2x^2-6x+5x-15}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{\left(2x+5\right)\left(x-3\right)}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)
Do đó: \(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}\)