Question

[Thùyy Lynhh]

\(\)A=\(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)với B=\(\dfrac{x-3}{x+1}\)

a) rút gọn A

b)  P=A.B,tìm x để P=\(\dfrac{9}{2}\)

c) tìm x để B<1

Answer 1

[Nguyễn Lê Phước Thịnh]

a: Ta có: \(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)

\(=\dfrac{2x^2-6x+x^2+4x+3-3+11x}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{3x^2+9x}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{3x}{x-3}\)

b: Ta có P=AB

nên \(P=\dfrac{3x}{x-3}\cdot\dfrac{x-3}{x+1}=\dfrac{3x}{x+1}\)

Để \(P=\dfrac{9}{2}\) thì 9x+9=6x

\(\Leftrightarrow3x=-9\)

hay x=-3(loại)

Answer 2

[ILoveMath]

a) \(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\\ \Rightarrow A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

    \(\Rightarrow A=\dfrac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

    \(\Rightarrow A=\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)-3+11x}{\left(x-3\right)\left(x+3\right)}\)

    \(\Rightarrow A=\dfrac{2x^2-6x+x^2+4x+3-3+11x}{\left(x-3\right)\left(x+3\right)}\)

    \(\Rightarrow A=\dfrac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)

    \(\Rightarrow A=\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

    \(\Rightarrow A=\dfrac{3x}{x-3}\)