Question

[Hieu Phan an]

Giải phương trình:

\(\frac{x-4}{x\left(x+2\right)}\) - \(\frac{1}{x\left(x-2\right)}\) = \(\frac{-2}{\left(x+2\right)\left(x-2\right)}\)

\(\frac{1}{x\left(x+3\right)}\) + \(\frac{1}{\left(x+3\right)\left(x+6\right)}\) + \(\frac{1}{\left(x+6\right)\left(x+9\right)}\) + \(\frac{1}{\left(x+9\right)\left(x+12\right)}\) = \(\frac{1}{16}\)

Answer 1

[Nguyễn Ngọc Lộc]

a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne\pm2\\x\ne0\end{matrix}\right.\)

Ta có : \(\frac{x-4}{x\left(x+2\right)}-\frac{1}{x\left(x-2\right)}=-\frac{2}{\left(x+2\right)\left(x-2\right)}\)

=> \(\frac{\left(x-4\right)\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}-\frac{x+2}{x\left(x-2\right)\left(x+2\right)}=-\frac{2x}{x\left(x+2\right)\left(x-2\right)}\)

=> \(\left(x-4\right)\left(x-2\right)-x-2=-2x\)

=> \(x^2-4x-2x+8-x-2=-2x\)

=> \(x^2-5x+6=0\)

=> \(\left(x-2\right)\left(x-3\right)=0\)

=> \(\left[{}\begin{matrix}x=2\\x=3\left(TM\right)\end{matrix}\right.\)

=> x = 3 .

Vậy phương trình trên có tập nghiệm là \(S=\left\{3\right\}\)

b, ĐKXĐ : \(x\ne0,-3,-6,-9,-12\)

Ta có : \(\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+12\right)}=\frac{1}{16}\)

=> \(\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}+\frac{1}{x+9}-\frac{1}{x+12}=\frac{1}{16}\)

=> \(\frac{1}{x}-\frac{1}{x+12}=\frac{1}{16}\)

=> \(\frac{x+12}{x\left(x+12\right)}-\frac{x}{x\left(x+12\right)}=\frac{1}{16}\)

=> \(x\left(x+12\right)=192\)

=> \(x^2+12x-192=0\)

=> \(x^2+2x.6+36-228=0\)

=> \(\left(x+6\right)^2=288\)

=> \(\left[{}\begin{matrix}x=\sqrt{288}-6\\x=-\sqrt{288}-6\end{matrix}\right.\) ( TM )

Vậy phương trình có tập nghiệm là \(S=\left\{\pm\sqrt{288}-6\right\}\)