Let $a,b,c,d,e$ be the roots of $x^5+x^4+2x^3+4x^2+7x+1=0$ the question is:find the value of $$\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2-c}+\frac{1}{2-d}+\frac{1}{2-e}$$ my try is in below
with respect to $2-a,2-b,...$ we can use $$(x+2)^5+(x+2)^4+2(x+2)^3+4(x+2)^2+7(x+2)+1=0$$ so we can find $$\frac{1}{a'}+\frac{1}{b'}+\frac 1{c'}+\frac{1}{d'}+\frac1{d'}$$
and it is standard for roots relation (I remember from the past)
$$x^5+\cdots+32+16+16+16+14+1=0\\(x-a')(x-b')(x-c')(x-d')(x-e')=0\\a'b'c'd'e'=-95$$and $$a'b'c'd'+a'b'c'e'+...=?\\x^180+x^132+x^116+x^124+7x=159x$$ so $$\frac{1}{a'}+\frac{1}{b'}+\frac 1{c'}+\frac{1}{d'}+\frac1{d'}=\frac{159}{95}$$
Am I right?
Is there a different technic to find the value?