Question

[Nguyễn Thị Thu Hằng]

Giải phương trình sau :

a, \(\dfrac{1}{x+1}+\dfrac{2}{x^3-x^2-x+1}+\dfrac{3}{x^2-1}=0\)

b, \(\dfrac{1}{x^2+5x+4}+\dfrac{1}{x^2+11x+28}+\dfrac{1}{x^2+17x+70}+\dfrac{1}{x^2+23x+130}=\dfrac{4}{13}\)

Answer 1

[Ánh Lê]

a)

\(\dfrac{1}{x+1}+\dfrac{2}{x^3-x^2-x+1}+\dfrac{3}{x^2-1}=0\) (\(x\ne\pm1\))

\(\Rightarrow\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)^2}+\dfrac{2}{\left(x+1\right)\left(x-1\right)^2}+\dfrac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)^2}=0\)

\(\Rightarrow\dfrac{x^2-2x+1+2+3x-3}{\left(x+1\right)\left(x-1\right)^2}=0\)

\(\Rightarrow\dfrac{x^2+x-2}{\left(x+1\right)\left(x-1\right)^2}=0\)

\(\Rightarrow x^2-x+2=0\)

\(\Rightarrow\left(x-1\right)\left(x+2\right)=0\)

=> Th1 :

x- 1 =0

=> x = 1 ( hư cấu vì không thỏa mãn ĐK )

Th2 :

x+2 = 0

=> x = -2 ( hợp lí )

Vậy nghiệm của phương trình là x = -2