ĐK:x≥1pt tương đương:3(√2x+1−1)22>2x+1+√x−13(2x+2−2√2x+1)2>2x+1+√x−1⇒{6x+6−6√2x+1>4x+2+2√x−1√2x+1(3√2x+1−3−2√2x+1)2>√x−1+1,5⇒{2x+1−6√2x+1>2√x−1−32x+1−3√2x+1>2√x−1+3⇒3√2x+1>6⇒x>2,5
ĐK:x≥1bpt tương đương:3(√2x+1−1)22>2x+1+√x−13(2x+2−2√2x+1)2>2x+1+√x−1⇒{6x+6−6√2x+1>4x+2+2√x−1√2x+1(3√2x+1−3−2√2x+1)2>√x−1+1,5⇒{2x+1−6√2x+1>2√x−1−32x+1−3√2x+1>2√x−1+3⇒3√2x+1>6⇒x>2,5