Đk: x≥1Ta có: 9x2−14x+25=(3x+3+4√2x−1)(3x+3−4√2x−1)Do đó :pt⟺x(3x+3−4√2x−1)=(√x−1−1)(2x−4)⟺3x2+3x−4x√2x−1=2√x−1(x−2)−2(x−2)⟺3x2+5x−4−4x√2x−1=2√x−1(x−2)⟺4x2−4x√2x−1+2x−1=x2−4x+4+2√x−1+x−1⟺(2x−√2x−1)2=(x−2+√x−1)2Đến đây bạn tự giải tiếp nhé!
Đk:
x≥1Ta có:
9x2−14x+25=(3x+3+4√2x−1)(3x+3−4√2x−1)Do đó :
pt⟺x(3x+3−4√2x−1)=(√x−1−1)(2x−4)⟺3x2+3x−4x√2x−1=2√x−1(x−2)−2(x−2)⟺3x2+5x−4−4x√2x−1=2√x−1(x−2)$\iff 4x^2-4x\sqrt{2x-1}+2x-1=x^2-4x+4+2\sqrt{x-1}
(x-2)+x-1$$\iff (2x-\sqrt{2x-1})^2=(x-2+\sqrt{x-1})^2$Đến đây bạn tự giải tiếp nhé!