[Toán 8]Cho: $\dfrac{35x-29}{{x}{2}-3x+2}=\dfrac{A}{x-1}+\dfrac{B}{x-2}$

T

thong7enghiaha



$\dfrac{35x-29}{{x}{2}-3x+2}=\dfrac{A}{x-1}+\dfrac{B}{x-2}$

\Leftrightarrow $\dfrac{35x-29}{(x-1)(x-2)}=\dfrac{(A+B)x-2A-B}{(x-1)(x-2)}$

Đồng nhất hệ số 2 vế ta có:

$A+B=35; -2A-B=-29$

\Rightarrow $A=-6; B=41$

\Rightarrow $A-B=-47$
 
O

ooeelnujikoo

Cho [TEX]\frac{35x-29}{{x}^{2}-3x+2}=\frac{A}{x-1}+\frac{B}{x-2}[/TEX] là 1 hằng đẳng thức . Tính A+B , A-B

[TEX]\text{quy dong }\\\ \frac{A}{x-1}+\frac{B}{x-2} = \frac{(A+B)x-2A-B}{x^2-3x+2}[/TEX]

[TEX]\left\{ \begin{array}{l}A+B=35\\ -2A-B=-29\end{array} \right.[/TEX]

[TEX]\left\{ \begin{array}{l}A=-6\\ B=41\end{aray} \right.[/TEX]

\Rightarrow [TEX]A+B=35, A-B=-47[/TEX]
 
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