Điều kiện: ⎩⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎧5x−1≥0⇔x≥513x−2≥0⇔x≥32x2−2x+5≥0∀xϵR
Đề bài suy ra :−2x−1+3x−2+5x−1+(x2−3x+2)x2−2x+5=0⇔(−2x+2)+(3x−2−1)+(5x−1−2)+(x2−2x−x+2)x2−2x+5=0⇔−2(x−1)+(x−1)(3x−2+13)+(x−1)(5x−1+25)+(x−1)(x−2)x2−2x+5=0⇔(x−1)(−2+3x−2+13+5x−1+25+(x−2)x2−2x+5)=0 ⇔(x−1)[(3x−2+13−1)+(5x−1+25−1)+(x−2)(x2−2x+5)]⇔(x−1)[(3x−2+13−3x−2−1)+(5x−1+25−5x−1−2)+(x−2)(x2−2x+5)]⇔(x−1)[(3x−2+12−3x−2)+(5x−1+23−5x−1)+(x−2)(x2−2x+5)]
Mình chia làm 2 trường hợp cho đỡ rối mắt
TH1: (3x−2+1).(3x−2+2)4−3x+2+(5x−1+2).(5x−1+3)9−5x+1+(x−2)x2−2x+5=0⇔(3x−2+1).(3x−2+2)6−3x+(5x−1+2).(5x−1+3)10−5x+(x−2)x2−2x+5=0⇔(x−2)((3x−2+1).(3x−2+2)−3+(5x−1+2).(5x−1+3)−5+x2−2x+5)=0
mà:x≥32(3x−2+1).(3x−2+2)−3+(5x−1+2).(5x−1+3)−5+x2−2x+5>0
Dấu đẳng thức không xãy ra. Chỉ còn x-2=0⇔ x=2(TMĐK)
TH2:x−1=0⇔x=1(TMĐK)
Vậy phương trình đã cho có 2 nghiệm x=1,x=2