Question

Các hệ số a,b thỏa mãn đẳng thức : \(\dfrac{1}{x^2-4}=\dfrac{a}{x-2}+\dfrac{b}{x+2}\)là

Answer 1

Ta co:

\(\dfrac{1}{x^2-4}=\dfrac{1}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow\dfrac{1}{\left(x-2\right)\left(x+2\right)}=\dfrac{a}{x-2}+\dfrac{b}{x+2}\)

\(\Rightarrow\dfrac{a\left(x+2\right)+b\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{ax+2a+bx-2b}{\left(x-2\right)\left(x+2\right)}\)

Answer 2

Ta có: \(\dfrac{1}{x^2-4}=\dfrac{a}{x-2}+\dfrac{b}{x+2}\Rightarrow\dfrac{1}{x^2-4}=\dfrac{ax+2a+bx-2b}{x^2-4}\)

\(\Rightarrow ax+2a+bx-2b=1\)

\(\Rightarrow x\left(a+b\right)+\left(2a-2b\right)=0x+1\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=0\\2a-2b=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{4}\\b=-\dfrac{1}{4}\end{matrix}\right.\)

Vậy: \(a=\dfrac{1}{4};b=-\dfrac{1}{4}\).