Question

Cho biểu thức:
P=(\(\dfrac{x }{x+3}\)-\(\dfrac{x^2+5}{x^2-9}+\dfrac{7}{x-3}\)) .\(\dfrac{x+3}{4}\)
a)Chứng minh P=\(\dfrac{x+4}{x-3}\)
b)Tính P tại x =5
c)Tìm x ∈ Z để P∈Z
-GIÚP MÌNH VỚI Ạ-

Answer 1

đkxđ: x\(\ne\pm3\)

a/ \(P=\left(\dfrac{x}{x+3}-\dfrac{x^2+5}{x^2-9}+\dfrac{7}{x-3}\right)\cdot\dfrac{x+3}{4}=\left(\dfrac{x\left(x-3\right)-x^2-5+7\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\cdot\dfrac{x+3}{4}=\dfrac{x^2-3x-x^2-5+7x+21}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{4}=\dfrac{4x+16}{x-3}\cdot\dfrac{1}{4}=\dfrac{4\left(x+4\right)}{4\left(x-3\right)}=\dfrac{x+4}{x-3}\)

b/ tại x = 5 thì:

\(P=\dfrac{5+4}{5-3}=\dfrac{9}{2}\)

c/ Ta có: \(\dfrac{x+4}{x-3}=\dfrac{x-3+7}{x-3}=\dfrac{x-3}{x-3}+\dfrac{7}{x-3}=1+\dfrac{7}{x-3}\)

để P ∈ Z thì \(\dfrac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)\)

=> x - 3 = {-7;-1;1;7}

=> x = {-4;2;4;10}

Vậy.............