If $\frac{3x+4}{x^2-3x+2}=\frac{A}{x-2}-\frac{B}{x-1}$ , then $\left(A,B\right)=$

$\frac{3x+4}{x^2-3x+2}=\frac{A\left(x-1\right)+B\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}$

$3x+4=Ax-A+Bx-2B$

$3x+4=\left(A+B\right)x+\left(-A-2B\right)$

$A+B=3\cdots \cdots \cdots \left(1\right)$

$\underset{\_}{-A-2B=4\cdots \cdots \left(2\right)}$

$-B=7\Rightarrow B=-7$

Now, substitute B value in eqn (1)

$A-7=3\Rightarrow A=10$

$\therefore \left(A,B\right)=\left(10,-7\right)$