Question

Cho hai phân thức \(P = \frac{{{x^2} + 6{\rm{x}} + 9}}{{{x^2} + 3{\rm{x}}}}\) và \(Q = \frac{{{x^2} + 3{\rm{x}}}}{{{x^2} - 9}}\)

a) Rút gọn P và Q

b) Sử dụng kết quả câu a, Tính P.Q và P:Q

Answer 1

\(a,P=\dfrac{x^2+6x+9}{x^2+3x}\\ =\dfrac{x^2+2\cdot3\cdot x+3^2}{x\left(x+3\right)}\\ =\dfrac{\left(x+3\right)^2}{x\left(x+3\right)}\\ =\dfrac{x+3}{x}\\ Q=\dfrac{x^2+3x}{x^2-9}\\ =\dfrac{x\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\\ =\dfrac{x}{x-3}\\ b,P\cdot Q=\dfrac{x+3}{x}\cdot\dfrac{x}{x-3}\\ =\dfrac{\left(x+3\right)\cdot x}{x\cdot\left(x-3\right)}\\ =\dfrac{x+3}{x-3}\\ P:Q=\dfrac{x+3}{x}:\dfrac{x}{x-3}\\ =\dfrac{x+3}{x}\cdot\dfrac{x-3}{x}\\ =\dfrac{x^2-9}{x^2}\)

 

Answer 2

a) P= \(\dfrac{x^2+6x+9}{x^2+3x}=\dfrac{\left(x+3\right)^2}{x\left(x+3\right)}=\dfrac{x+3}{x}\)
    Q= \(\dfrac{x^2+3x}{x^2-9}=\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x}{x-3}\)
b)\(P.Q=\dfrac{x+3}{x}.\dfrac{x}{x-3}=1\)
  \(P:Q=\dfrac{x+3}{x}:\dfrac{x}{x-3}=\dfrac{x+3}{x}.\dfrac{x-3}{x}=\dfrac{x^2-9}{x^2}\)