Question

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

Rút gọn B và tìm x để B=1

Answer 1

đk: x #4; x > 0

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

\(=\dfrac{\left(x-\sqrt{x}\right)-2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-1=\dfrac{x-3\sqrt{x}-4}{x-4}-\dfrac{x-4}{x-4}=\dfrac{-3\sqrt{x}}{x-4}\)

B = 1

<=> \(B=\dfrac{-3\sqrt{x}}{x-4}=1\Leftrightarrow-3\sqrt{x}=x-4\)

<=> x = 1