Hoặc
ĐKXĐ: 2x+1≥0⇒x≥−12 .
x−42=2x+1⇔x−42=2x+12⇔x2−8x+16=4x2+4x+1⇔3x2+12x−15=0⇔x2+4x−5=0⇔x2+5x−x−5=0⇔x+5x−1=0⇔x+5=0x−1=0⇔x=−5 ktmx=1 tm
Vậy x=1 .