Đáp án:
a)2x−15x−10=0DK5x−10≠0⇔x≠2⇔2x−1=0⇔x=122x−15x−10=0DK5x−10≠0⇔x≠2⇔2x−1=0⇔x=12
b)x2−x2x=0DKx≠0⇔x(x−1)2x=0⇔(x−1)2=0⇔x−1=0⇔x=1x2−x2x=0DKx≠0⇔x(x−1)2x=0⇔(x−1)2=0⇔x−1=0⇔x=1
c)2x+34x−5=0DK4x−5≠0⇔≠54⇒2x+3=0⇔x=−322x+34x−5=0DK4x−5≠0⇔≠54⇒2x+3=0⇔x=−32(nhận)
d)(x−1)(x+2)x2−4x+3=0DKx2−4x+3≠0⇔x2−3x−x+3≠0⇔(x−3)(x−1)≠0⇔x−3≠0,x−1≠0⇔x≠3,x≠1⇒(x−1)(x+2)(x−3)(x−1)=0⇔(x+2)(x−3=0⇔x+2=0⇔x=−2(x−1)(x+2)x2−4x+3=0DKx2−4x+3≠0⇔x2−3x−x+3≠0⇔(x−3)(x−1)≠0⇔x−3≠0,x−1≠0⇔x≠3,x≠1⇒(x−1)(x+2)(x−3)(x−1)=0⇔(x+2)(x−3=0⇔x+2=0⇔x=−2(nhận)
e)x2−1x2−2x+1=0DKx2−2x+1≠0⇔(x−1)2≠0⇔x−1≠0⇔x≠1⇒(x−1)(x+1)(x−1)2=0⇔(x+1)(x−1)=0⇔x+1=0⇔x=−1
Giải thích các bước giải:
