Question

\(A=\frac{x+2}{x\sqrt{x}+1}+\frac{\sqrt{x}-1}{x-\sqrt{x}+1}-\frac{\sqrt{x}-1}{x-1}\)

a, Rút gọn A (x\(x\ge0,x\ne1\))

b, Tìm giá trị của A khi x=4

Answer 1

a.

\(A=\frac{x+2}{\sqrt{x^3}+1}+\frac{\sqrt{x}-1}{x-\sqrt{x}+1}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ =\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\\ =\frac{x+2+x-1-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\frac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\\ =\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\\ =\frac{\sqrt{x}}{x-\sqrt{x}+1}\)

b. Khi x=4 ta có:

\(A=\frac{\sqrt{4}}{4-\sqrt{4}+1}=\frac{2}{4-2+1}=\frac{2}{3}\)