Question

rút gon các biểu thức sau

a) \(A=\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}+\dfrac{4-x}{\sqrt{x}-2}\) với x ≥0, x ≠4

Answer 1

\(A=\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}+\dfrac{4-x}{\sqrt{x}-2}=\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}+2}+\dfrac{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-2}=\sqrt{x}+2-\sqrt{x}-2=0\)

Answer 2

Ta có: \(A=\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}+2}+\dfrac{4-x}{\sqrt{x}-2}\)

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

= \(\dfrac{\left(\sqrt{x}+2\right)\left(x-4+4-x\right)}{x-4}=\dfrac{\left(\sqrt{x}+2\right).0}{x-4}=0\)

Vậy A=0