Question

Cho bieu thuc:P=\(\dfrac{\left(a+3\right)^2}{a^2+3a}\times\left(1-\dfrac{6a-18}{a^2-9}\right)\)voi a ≠0;a≠ +-3

a)rut gon bieu thuc P

b)tim gia tri cua a de P= -2

c)tim cac gia tri nguyen cua a de bieu thuc P co gia tri nguyen

mng giup minh voi mai thi rui!

Answer 1

a: \(P=\dfrac{a+3}{a}\cdot\dfrac{a^2-9-6a+18}{\left(a-3\right)\left(a+3\right)}\)

\(=\dfrac{\left(a-3\right)^2}{a\left(a-3\right)}=\dfrac{a-3}{a}\)

b: Để P=-2 thì -2a=a-3

=>-3a=-3

=>a=1

c: Để P nguyên thì a-3 chia hết cho a

=>-3 chia hết cho a

mà a<>0; a<>3; a<>-3

nên \(a\in\left\{1;-1\right\}\)