Question

[co gai buong binh]

a) \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)

b) \(\dfrac{a+2b}{3a-b}+\dfrac{2a-5b}{b-3a}\)

c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)

d) \(\dfrac{4x}{x^2-4}+\dfrac{x}{x+2}+\dfrac{2}{x-2}\)

e) \(\dfrac{3x^2-x+3}{x^3-1}+\dfrac{1-x}{x^2+x+1}+\dfrac{2}{1-x}\)

f) \(\dfrac{1}{x^2+3x+2}+\dfrac{1-x}{x^2+x+1}+\dfrac{2}{1-x}\)

g) \(\dfrac{a^3}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^3}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^3}{\left(c-a\right)\left(c-b\right)}\)

h) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}\)

Answer 1

[Nguyễn Ngọc Linh]

a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)

\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)

Answer 2

[Nguyễn Ngọc Linh]

c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)

Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)

\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)