Question

Cho biểu thức Q= (\(\dfrac{4\sqrt{x}}{x+\sqrt{x}}+\dfrac{8}{2-\sqrt{x}}\)) : (\(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\))

Tìm giá trị của x khi Q=-1

Answer 1

\(Q=\left(\dfrac{4\sqrt{x}}{x+2\sqrt{x}}+\dfrac{8}{2-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\dfrac{4x-8\sqrt{x}-8x-16\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-1-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{-4x-20\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}-1}\)

\(=\dfrac{4x+20\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)

Để Q=-1 thì \(4x+20\sqrt{x}=-x-3\sqrt{x}-2\)

=>\(5x+23\sqrt{x}+2=0\)

hay \(x\in\varnothing\)