\(\dfrac{4x-7}{x^2-3x+2}\)=\(\dfrac{a}{x-1}\)+\(\dfrac{b}{x-2}\)
\(\Leftrightarrow\) \(\dfrac{4x-7}{x^2-3x+2}\)=\(\dfrac{a\left(x-2\right)+b\left(x-1\right)}{\left(x-1\right).\left(x-2\right)}\)
\(\Leftrightarrow\)\(\dfrac{4x-7}{x^2-3x+2}\)=\(\dfrac{a\left(x-2\right)+b\left(x-1\right)}{x^2-3x+2}\)
\(\Rightarrow\) 4x - 7 = a(x-2)+b(x-1)
\(\Rightarrow\) 4x - 7 = ax - 2a + bx - b
\(\Rightarrow\) 4x - 7 = x(a+b) - (2a + b) (1)
Từ pt(1) \(\Rightarrow\) \(\begin{cases} a+b=4\\ 2a+b=7 \end{cases}\)
\(\Leftrightarrow\) \(\begin{cases} 2a+2b=8\\ 2a+b=7 \end{cases}\)
\(\Leftrightarrow\) b=1
\(\Rightarrow\) 2a + 1=7
\(\Rightarrow\)a=(7-1):2=3
Vậy a=3, b=1 .
Đúng thì tick cho mình nha.
