Question

\(A=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\frac{\sqrt{x}-3}{2\sqrt{x}-x}\)

a, Rút gọn

b, tìm x để A>0

c, tìm x để A=1

Answer 1

a)\(A=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\frac{\sqrt{x}-3}{2\sqrt{x}-x}\)

\(A=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}+\frac{4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right).\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)

\(A=\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)

\(A=\frac{x+4\sqrt{x}+4-\left(x-4\sqrt{x}+4\right)+4x}{2+\sqrt{x}}.\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(A=\frac{x+4\sqrt{x}+4-x+4\sqrt{x}-4+4x}{2+\sqrt{x}}.\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(A=\frac{8\sqrt{x}+4x}{2+\sqrt{x}}.\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(A=\frac{4\sqrt{x}(2+\sqrt{x})}{2+\sqrt{x}}.\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(A=\frac{4x}{\sqrt{x}-3}\)

Answer 2

đk: \(x\ne4;x\ne9;x\ge0\)

\(A>0\)\(=>A=\frac{4x}{\sqrt{x}-3}>0\)

có \(x\ge0=>4x\ge0\)

\(=>A>0=>\sqrt{x}-3>0< =>\sqrt{x}>3< =>x>9\)

vậy x>9 thì A>0