(1/x^2-3x+2)+(1/x^2-5x+6)+(1/x^2-7x+12)+(1/x^2-9x+20)=1/15

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Solution for (1/x^2-3x+2)+(1/x^2-5x+6)+(1/x^2-7x+12)+(1/x^2-9x+20)=1/15 equation: 



(1/x^2-3x+2)+(1/x^2-5x+6)+(1/x^2-7x+12)+(1/x^2-9x+20)=1/15

We move all terms to the left:

(1/x^2-3x+2)+(1/x^2-5x+6)+(1/x^2-7x+12)+(1/x^2-9x+20)-(1/15)=0



Domain of the equation: x^2-3x+2)!=0

x∈R





Domain of the equation: x^2-5x+6)!=0

x∈R





Domain of the equation: x^2-7x+12)!=0

x∈R





Domain of the equation: x^2-9x+20)!=0

x∈R



We add all the numbers together, and all the variables

(1/x^2-3x+2)+(1/x^2-5x+6)+(1/x^2-7x+12)+(1/x^2-9x+20)-(+1/15)=0

We get rid of parentheses

1/x^2-3x+1/x^2-5x+1/x^2-7x+1/x^2-9x+2+6+12+20-1/15=0

We calculate fractions

There is no solution for this equation

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