Question

Giải phương trình:

\(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}=\frac{1}{8}\)

Answer 1

\(\Leftrightarrow\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x-1}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-3}+...+\frac{1}{x-4}-\frac{1}{x-5}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x-1}-\frac{1}{x-5}=\frac{1}{8}\)

\(\Leftrightarrow\frac{x-5-x+1}{\left(x-1\right)\left(x-5\right)}=\frac{1}{8}\)

\(\Leftrightarrow-4.8=x^2-6x+5\)

\(\Leftrightarrow x^2-6x+37=0\)

Answer 2

bo tay

Answer 3

kb với tôi đi

Answer 4

\(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}=\frac{1}{8}\)

\(\Rightarrow\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}=\frac{1}{8}\)

\(\Rightarrow\frac{1}{x-1}-\frac{1}{x-2}+...+\frac{1}{x-4}-\frac{1}{x-5}=\frac{1}{8}\)

\(\Rightarrow\frac{1}{x-1}-\frac{1}{x-5}=\frac{1}{8}\)

\(...\)