3)x+1+x2−4x+1=3x" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">x+1+x2−4x+1−−−−−−−−−−√=3x√x+1+x2−4x+1=3x⇔(x+1)+(x−1)2−6x=3x" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">⇔(x+1)+(x+1)2−6x−−−−−−−−−−−√=3x√x⇔(x+1)+(x−1)2−6x=3x⇔1+1−6(xx+1)2=3(xx+1)" role="presentation" style="font-size: 13.696px; position: relative;">ta chia ca hai ve cho x+1#0 ,có⇔1+1−6(x√x+1)2−−−−−−−−−−−√=3(x√x+1)⇔1+1−6(xx+1)2=3(xx+1)Đặt xx+1=a" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">x√x+1=axx+1=a→" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">→→ PT trở thành: 1+1−6a2=3a" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">1+1−6a2−−−−−−√=3a1+1−6a2=3a→..................." role="presentation" style="font-size: 13.696px; display: inline; position: relative;">→...................