If $\frac{x^2-5x+7}{{\left(x-1\right)}^4}=\frac{A}{x-1}+\frac{B}{{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^3}+\frac{D}{{\left(x-1\right)}^4}$  , then $A+B+C+D=$

$\frac{2^2-5\left(2\right)+7}{{\left(2-1\right)}^2}=A+B+C+D$

$11-10=A+B+C+D$

$\therefore A+B+C+D=1$

(or)

$\frac{x^2-5x+7}{{\left(x-1\right)}^4}$  (let $x=-1=y\Rightarrow x=y+1$ )

$\Rightarrow \frac{y^2=2y+1-5y-5+7}{y^4}=\frac{1}{y^2}-\frac{3}{y^3}+\frac{3}{y^4}$

$A+B+C+D=0+1-3+3=1$