If $\frac{x^4}{\left(x-1\right)\left(x+2\right)}=x^2+3x+7+\frac{A}{x-2}+\frac{B}{x-1}$  then

$x^2-3x+2)x^4(x^2+3x+7$

$\underset{\_}{x^4-3x^3+2x^2}$

$3x^3-2x^2$

$\underset{\_}{3x^3-9x^2+6x}$

$7x^2-6x$

$\underset{\_}{7x^2-21x+14}$

$\underset{\_}{15x-14}$

$\frac{x^4}{\left(x-1\right)\left(x+2\right)}=x^2+3x+7+\frac{15x-14}{x^2-3x+2}$

$\frac{15x-14}{x^2-3x+2}=\frac{A}{x-2}+\frac{B}{x-1}$

$15x-14=A\left(x-1\right)+B\left(x-2\right)$

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

$A+B=15\cdots \cdots \cdots \left(1\right)$

$\underset{\_}{-A-2B=-14}\cdots \cdots \left(2\right)$

$-B=1\Rightarrow B=-1$

$\text{Eqn}\left(1\right)\Rightarrow A-1=15$

$A=15+1$

$A=16$