Question

[30.Nɠυұễɳ Tɦàɲɦ Pɦúƈ ヅ]

Bài 4: Tính và rút gọn

C = \(\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}\) 

D = \(\dfrac{1}{x-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}+\dfrac{1}{x+\sqrt{x}}\:\:\left(x\ne1;x>0\right)\)

Answer 1

[Nguyễn Lê Phước Thịnh]

\(C=\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}\)

\(=\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{9-5}}\)

\(=\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{6+2\sqrt{5}}{4}}\)

\(=\dfrac{2\left(\sqrt{5}-1\right)}{5-1}+\dfrac{\sqrt{6+2\sqrt{5}}}{2}\)

\(=\dfrac{\sqrt{5}-1}{2}+\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{2}\)

\(=\dfrac{\sqrt{5}-1}{2}+\dfrac{\sqrt{5}+1}{2}=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{2}=\dfrac{2\sqrt{5}}{2}=\sqrt{5}\)

\(D=\dfrac{1}{x-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}+\dfrac{1}{x+\sqrt{x}}\)

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

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

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

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