Question

Answer 1

a: ĐKXĐ: x<>3

\(x^2-9=0\)

=>\(x^2=9\)

=>x=3(loại) hoặc x=-3(nhận)

Thay x=-3 vào A, ta được:

\(A=\frac{-3+2}{-3-3}=\frac{-1}{-6}=\frac16\)

b: ĐKXĐ: x<>-2 x<>1

\(B=\frac{1}{x+2}-\frac{x}{1-x}-\frac{3x}{x^2+x-2}\)

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

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

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

c: P=A*B

\(=\frac{x+2}{x-3}\cdot\frac{x+1}{x+2}=\frac{x+1}{x-3}\)

\(=\frac{x-3+4}{x-3}=1+\frac{4}{x-3}\)

Để P có giá trị nguyên thì 4⋮x-3

=>x-3∈{1;-1;2;-2;4;-4}

=>x∈{4;2;5;1;7;-1}

Kết hợp ĐKXĐ, ta được: x∈{4;5;7;-1}