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