Question

Answer 1

ĐKXĐ: x∉{0;-1;-2;-3;-4;-5;-6;-7}

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

=>\(\frac{x+7+x}{x\left(x+7\right)}+\frac{x+2+x+5}{\left(x+2\right)\left(x+5\right)}=\frac{x+4+x+3}{\left(x+4\right)\left(x+3\right)}+\frac{x+6+x+1}{\left(x+6\right)\left(x+1\right)}\)

=>\(\frac{2x+7}{x\left(x+7\right)}+\frac{2x+7}{\left(x+2\right)\left(x+5\right)}=\frac{2x+7}{\left(x+4\right)\left(x+3\right)}+\frac{2x+7}{\left(x+6\right)\left(x+1\right)}\)

=>\(\left(2x+7\right)\left(\frac{1}{x\left(x+7\right)}+\frac{1}{\left(x+2\right)\left(x+5\right)}-\frac{1}{\left(x+4\right)\left(x+3\right)}-\frac{1}{\left(x+6\right)\left(x+1\right)}\right)=0\)

=>2x+7=0

=>2x=-7

=>\(x=-\frac72\) (nhận)