Question

Answer 1

a: Thay x=3 vào B, ta được:

\(B=\frac{3-3}{3+1}=\frac04=0\)

b: ĐKXĐ: x<>3; x<>-3

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

\(=\frac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)+11x-3}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{2x^2-6x+x^2+4x+3+11x-3}{\left(x-3\right)\left(x+3\right)}=\frac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3x}{x-3}\)

c: P=A*B

\(=\frac{3x}{x-3}\cdot\frac{x-3}{x+1}=\frac{3x}{x+1}\)

Để P là số nguyên thì 3x⋮x+1

=>3x+3-3⋮x+1

=>-3⋮x+1

=>x+1∈{1;-1;3;-3}

=>x∈{0;-2;2;-4}