Question

Giúp tôi với! help me

Answer 1

a: ĐKXĐ: x∉{3;-3}

\(P=\frac{x^2-6x+9}{9-x^2}+\frac{4x+8}{x+3}\)

\(=\frac{\left(x-3\right)^2}{\left(3-x\right)\left(3+x\right)}+\frac{4x+8}{x+3}=\frac{3-x}{3+x}+\frac{4x+8}{x+3}=\frac{4x+8+3-x}{x+3}=\frac{3x+11}{x+3}\)

b: Khi x=7 thì \(P=\frac{3\cdot7+11}{7+3}=\frac{33+11}{10}=\frac{44}{10}=4,4\)

c: \(P=\frac{3x+11}{x+3}=\frac{3x+9+2}{x+3}=3+\frac{2}{x+3}\)

Để P là số nguyên thì 2⋮ x+3

=>x+3∈{1;-1;2;-2}

=>x∈{-2;-4;-1;-5}

Answer 2

có ai giúp ko