Question

cho biểu thức \(P=\left(\dfrac{3\sqrt{x}-1}{x-1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{1}{x+\sqrt{x}}\) với x>0 và x≠1. tìm x để 2P - x = 3

Answer 1

\(P=\left(\dfrac{3\sqrt{x}-1}{x-1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{1}{x+\sqrt{x}}\)

\(P=\left(\dfrac{3\sqrt{x}-1-\left(\sqrt{x}+1\right)}{x-1}\right):\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(P=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\left[\sqrt{x}\left(\sqrt{x}+1\right)\right]\)

\(P=\dfrac{2x}{\sqrt{x}-1}\)

ta có 2P-x=3

⇔\(2\cdot\dfrac{2x}{\sqrt{x}-1}-x=3\)

⇔\(4x-x\left(\sqrt{x}-1\right)=3\left(\sqrt{x}-1\right)\)

⇔\(5x-x\sqrt{x}-3\sqrt{x}+3=0\)