2√x+3+3√2x−3−√2x−2=√2(x+1)
ĐKXĐ : {2x−3≥02x−2≥0x+3+3√2x−3≥0
Chia 2 vế cho √2
PT⇔√2x+6+2.3√2x−3−√x−1=x+1
⇔√2x−3+2.3√2x−3+9−√x−1=x+1
⇔√(√2x−3+3)2−√x−1=x+1
⇔√2x−3+3−√x−1=x+1
⇔√2x−3−√x−1=x−2
⇔x−2√2x−3+√x−1=x−2
⇔(x−2)(1−1√2x−3+√x−1)=0
TH1:x−2=0⇔x=2
TH2:√2x−3+√x−1=1
⇔1−√2x−3=√x−1
⇒2x−2−2√2x−3=x−1
⇔x−1=2√2x−3
⇒x2−2x+1=4(2x−3)=8x−12
⇔x2−10x+13=0
⇔x2−10x+25−12=0
⇔(x−5)2−12=0⇔(x−5−√12)(x−5+√12)=0
⇔x=±√12+5
Thử lại : các nghiệm trên thoả PT
Vậy x=2;±√12+5