Câu 1: (x+4)2−6√x3+3x=13 (ĐKXĐ: x>0)<=>(x+4)2−13=6√x3+3x
<=>(x+4)4−26(x+4)2+169=36(x3+3x)
<=>(x4+16x3+96x2+256x+256)−(26x2+208x+416)+169=36x3+108x
<=>x4−20x3+70x2−60x+9=0
<=>(x−1)(x−3)(x2−16x+3)=0
<=>x=1 hoặc x=3 hoặc x=8±√61 (TMĐK)
Câu 2: 23√x−2+5√x+1−12=0 (ĐKXĐ: x > -1)
<=>(23√x−2−2)+(5√x+1−10)=0
<=>8(x−2)−843√(x−2)2+43√x−2+4+5(x+1−4)√x+1+2=0
<=>8(x−3)43√(x−2)2+43√x−2+4+5(x−3)√x+1+2=0
<=>(x−3)(843√(x−2)2+43√x−2+4+5√x+1+2)=0 (*)
Vì 843√(x−2)2+43√x−2+4+5√x+1+2>0 Với mọi x > -1
Nên (*) <=> x−3=0 <=>x=3