2 câu trả lời
ĐKXĐ:x≠1;3
x+1x−1+2x−5x−3=3
⇔(x+1)(x−3)(x−3)(x−1)+(2x−5)(x−1)(x−1)(x−3)=3
⇔x2−2x−3(x−3)(x−1)+2x2−7x+5(x−1)(x−3)=3
⇔x2−2x−3+2x2−7x+5(x−3)(x−1)=3
⇔3x2−9x+2(x−3)(x−1)=3
⇔3x2−9x+2=3(x−3)(x−1)
⇔3x2−9x+2=3(x2−4x+3)
⇔3x2−9x+2=3x2−12x+9
⇔−9x+2=−12x+9
⇔3x=7
⇔x=73
Vậy S={73}
x+1x-1+2x-5x-3=3(1)
ĐKXĐ:x≠1,x≠3
(1)⇔(x+1)(x-3)(x-1)(x-3)+(2x-5)(x-1)(x-3)(x-1)=3(x-3)(x-1)(x-3)(x-1)
⇒(x+1)(x-3)+(2x-5)(x-1)=3(x-3)(x-1)
⇔x2-2x-3+2x2-7x+5=3(x2-4x+3)
⇔3x2-9x+2=3x2-12x+9
⇔3x2-9x+2-3x2+12x-9=0
⇔3x-7=0
⇔3x=7
⇔x=73 (t/m)
Vậy x=73 là nghiệm của phương trình.