Question

[Trần gia hào]

1/(x-1)(x-2)+1/(x-2)(x-3)+1/(x-3)(x-4)+1/(x-4)(x-5)+1/(x-5)(x-6)=1/10

Answer 1

[Hạ Vy]

Ta có: \(\frac{1}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

Lại Có: ĐKXĐ: x≠1,x≠2,x≠3,x≠4,x≠5,x≠6

\(\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{10}\)<=>\(\frac{1}{\left(x-6\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-1\right)}=\frac{1}{10}\)

<=>\(\frac{1}{x-6}-\frac{1}{x-5}+\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-4}-\frac{1}{x-3}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-1}=\frac{1}{10}\)

<=> \(\frac{1}{x-6}-\frac{1}{x-1}=\frac{1}{10}\)

<=> \(\frac{x-1-x+6}{\left(x-6\right)\left(x-1\right)}=\frac{1}{10}\)

<=> \(\frac{5}{\left(x-6\right)\left(x-1\right)}=\frac{1}{10}\)

<=>(x-6)(x-1)=50

<=>x²-7x+6-50=0

<=>x²+4x-11x-44=0

<=>x(x+4)-11(x+4)=0

<=>(x+4)(x-11)=0

<=>\(\left[{}\begin{matrix}x+4=0\\x-11=0\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x=-4\\x=11\end{matrix}\right.\)_{(Thỏa mãn)}

_{Vậy phương trình thuộc tập nghiệm S={-4;11}}