Question

 Giai phuong trinh

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

Answer 1

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

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

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

<=>\(\frac{1}{x^2+1}-\frac{1}{x^2+5}=-1\)

<=>(x²+5)-(x²+1)=-(x²+1)(x²+5)

<=>4=-x⁴-6x²-5

<=>x⁴+6x²+9=0

<=>(x²+3)²=0

<=>x²+3=0

Do x²>0

=>x²+3>0 nên PT vô nghiệm