Question

 giúp em câu 7 với

Answer 1

a: ĐKXĐ: x∉{3;-3}

Ta có: \(A=\frac{x}{x-3}+\frac{4}{x+3}-\frac{x-21}{x^2-9}\)

\(=\frac{x}{x-3}+\frac{4}{x+3}-\frac{x-21}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x\left(x+3\right)+4\left(x-3\right)-x+21}{\left(x-3\right)\left(x+3\right)}=\frac{x^2+3x+4x-12-x+21}{\left(x-3\right)\left(x+3\right)}\)

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

b: Khi x=-6 thì \(A=\frac{-6+3}{-6-3}=\frac{-3}{-9}=\frac13\)

c: Để A nguyên thì x+3⋮x-3

=>x-3+6⋮x-3

=>6⋮x-3

=>x-3∈{1;-1;2;-2;3;-3;6;-6}

=>x∈{4;2;5;1;6;0;9;-3}

Kết hợp ĐKXĐ, ta được: x∈{4;2;5;1;6;0;9}