Giải phương trình (9x^2-4).(x+1)=(3x+2).(x^2-1 18/07/2021 Bởi Maya Giải phương trình (9x^2-4).(x+1)=(3x+2).(x^2-1
(9x²-4)(x+1)=(3x+2)(x²-1) ⇔(9x²-4)(x+1)-(3x+2)(x-1)(x+1)=0 ⇔(x+1)[9x²-4-(3x+2)(x-1)]=0 ⇔(x+1)[(3x-2)(3x+2)-(3x+2)(x-1)]=0 ⇔(x+1)(3x+2)(3x-2-x+1)=0 ⇔(x+1)(3x+2)(2x-1)=0 ⇔x+1=0⇒x=-1 3x+2=0⇒3x=-2⇒x=-2/3 2x-1=0⇒2x=1⇒x=1/2 Vậy S={-1;-1/3;1/2} Bình luận
Đáp án:
Giải thích các bước giải:
(9x²-4)(x+1)=(3x+2)(x²-1)
⇔(9x²-4)(x+1)-(3x+2)(x-1)(x+1)=0
⇔(x+1)[9x²-4-(3x+2)(x-1)]=0
⇔(x+1)[(3x-2)(3x+2)-(3x+2)(x-1)]=0
⇔(x+1)(3x+2)(3x-2-x+1)=0
⇔(x+1)(3x+2)(2x-1)=0
⇔x+1=0⇒x=-1
3x+2=0⇒3x=-2⇒x=-2/3
2x-1=0⇒2x=1⇒x=1/2
Vậy S={-1;-1/3;1/2}