a) 9−12x+4x2=4+x⇔(3−2x)2=4+x
⇔∣3−2x∣=4+x
th1: 3−2x≥0⇔2x≤3⇔⇔x≤23
⇒∣3−2x∣=4+x⇔3−2x=4+x⇔3x=−1⇔x=3−1(tmđk)
th2: 3−2x<0⇔2x>3⇔x>23
⇒∣3−2x∣=4+x⇔2x−3=4+x⇔x=7(tmđk)
vậy x=3−1;x=7
b) 4−4x+x2=(x−1)2+x−6
⇔(2−x)2=x2−2x+1+x−6
⇔∣2−x∣=x2−x−5
th1: 2−x≥0⇔x≤2
⇒∣2−x∣=x2−x−5⇔2−x=x2−x−5
⇔x2=7⇔{x=7(loại)x=−7(tmđk)
th2: 2−x<0⇔x>2
⇒∣2−x∣=x2−x−5⇔x−2=x2−x−5
⇔x2−2x−3=0⇔x2+x−3x−3=0
⇔x(x+1)−3(x+1)=0⇔(x−3)(x+1)=0
⇔{x−3=0x+1=0⇔{x=3(tmđk)x=−1(loại)
vậy x=−7;x=3