Question

giải phương trình:

`1/(x^2+x) + 1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)=5/(x+4)`

 

Answer 1

ĐKXĐ: x∉{0;-1;-2;-3;-4}

Ta có:\(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}=\frac{5}{x+4}\)

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

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

=>\(\frac{1}{x}-\frac{1}{x+4}=\frac{5}{x+4}\)

=>\(\frac{1}{x}=\frac{6}{x+4}\)

=>6x=x+4

=>5x=4

=>\(x=\frac45\) (nhận)