Question

Tìm các số a và b sao cho \(\frac{17x+18}{3x^2+x-14}=\frac{a}{x-2}+\frac{b}{3x+7}\)

Answer 1

\(\frac{17x+18}{3x^2+x-14}=\frac{a}{x-2}+\frac{b}{3x+7}\)

\(\Rightarrow\frac{17x+18}{3x^2+x-14}=\frac{a\left(3x+7\right)+b\left(x-2\right)}{\left(x-2\right)\left(3x+7\right)}\)

\(\Rightarrow\frac{17x+18}{3x^2+x-14}=\frac{3ax+7a+bx-2b}{3x^2+x-14}\)

\(\Rightarrow\frac{17x+18}{3x^2+x-14}=\frac{3ax+5a+bx}{3x^2+x-14}\)

\(\Rightarrow\frac{17x+18}{3x^2+x-14}=\frac{\left(3a+b\right)x+5a}{3x^2+x-14}\)

Đồng nhất hệ số, ta có: \(\hept{\begin{cases}3a+b=17\\5a=18\end{cases}}\Leftrightarrow\hept{\begin{cases}b=\frac{31}{5}\\a=\frac{18}{5}\end{cases}}\)