Question

Rút gọn A=\(\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}+1}+\dfrac{4}{x-1}\)

Answer 1

\(A=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}+1}+\dfrac{4}{x-1}\\ =\dfrac{\sqrt{x}+3+5}{\sqrt{x}+1}+\dfrac{4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+8\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{x-7\sqrt{x}-8+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{x-7\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

Answer 2

đk : x >= 0 ; x khác 1 

\(A=\dfrac{x+2\sqrt{x}-3+5\sqrt{x}-5+4}{x-1}=\dfrac{x+7\sqrt{x}-4}{x-1}\)