Giải phương trình sau:
a. \(\dfrac{6x+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
b. \(\dfrac{2}{x^2-4}-\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x-4}{x\left(x+2\right)}=0\)
c. \(\dfrac{1}{3-x}-\dfrac{1}{x+1}=\dfrac{x}{x-3}-\dfrac{\left(x-1\right)^2}{x^2-2x-3}\)
d. \(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)
e. \(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x-3}\)
f. \(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{x^6-1}\)
m.n jup vs mai kt r
Bài này t mới làm xong :v
d) \(\dfrac{1}{x-2}\) - \(\dfrac{6}{x+3}\) = \(\dfrac{5}{6-x^2-x}\) (ĐKXĐ : x \(\ne\) 2; x \(\ne\) -3)
<=> \(\dfrac{1}{x-2}\) - \(\dfrac{6}{x+3}\) = \(\dfrac{-5}{\left(x-2\right)\left(x+3\right)}\)
=> (x + 3) - 6(x - 2) = -5
<=> x + 3 - 6x + 12 = -5
<=> -5x + 15 = -5
<=> 5x = 20
<=> x = 4
Vậy tập nghiệm của PT là : S = { 4 }
