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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)
a) \(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\\ \Rightarrow A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow A=\dfrac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow A=\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)-3+11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow A=\dfrac{2x^2-6x+x^2+4x+3-3+11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow A=\dfrac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow A=\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow A=\dfrac{3x}{x-3}\)
\(3x^3-7x^2+17x-5=3x^3-x^2-6x^2+2x+15x-5\)
\(=x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)
\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
\(x^3-x^2-4=x^3+x^2+2x-2x^2-2x-4\)
\(=x\left(x^2+x+2\right)=2\left(x^2+x+2\right)=\left(x-2\right)\left(x^2+x+2\right)\)
Ta có: \(2x^4+11x^3+11x^2-24x-36\)
\(=2x^4+4x^3+7x^3+14x^2-3x^2-6x-18x-36\)
\(=\left(x+2\right)\left(2x^3+7x^2-3x-18\right)\)
\(=\left(x+2\right)\left(2x^3+4x^2+3x^2+6x-9x-18\right)\)
\(=\left(x+2\right)^2\left(2x^2+3x-9\right)\)
\(=\left(x+2\right)^2\cdot\left(2x^2+6x-3x-9\right)\)
\(=\left(x+2\right)^2\cdot\left(x+3\right)\left(2x-3\right)\)
Ta có: \(x^5+7x^4+21x^3+47x^2+80x+60\)
\(=x^5+4x^4+4x^3+3x^4+12x^3+12x^2+5x^3+20x^2+20x+15x^2+60x+60\)
\(=\left(x^2+4x+4\right)\left(x^3+3x^2+5x+15\right)\)
\(=\left(x+2\right)^2\cdot\left(x+3\right)\left(x^2+5\right)\)
Ta có: \(\frac{2x^4+11x^3+11x^2-24x-36}{x^5+7x^4+21x^3+47x^2+80x+60}\)
\(=\frac{\left(x+2\right)^2\cdot\left(x+3\right)\left(2x-3\right)}{\left(x+2\right)^2\cdot\left(x+3\right)\left(x^2+5\right)}=\frac{2x-3}{x^2+5}\)