Question

[Nguyễn Anh Thư]

  Bài 3
Cho A= \(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\)
a) Tìm ĐKXĐ của A
b) Chứng minh A= \(\frac{\sqrt{x}+1}{\sqrt{x}}\)
c) Tính A khi x=\(\frac{1}{4}\); x= \(3+2\sqrt{2}\)

Answer 1

[Nguyễn Minh Quang]

a. ĐKXĐ: 

\(\hept{\begin{cases}\sqrt{x}-1\ne0\\x-\sqrt{x}\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x>0\\x\ne1\end{cases}}}\)

b. ta có \(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-1}{\sqrt{x}.\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)

c. khi \(x=\frac{1}{4}\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow A=\frac{\frac{1}{2}+1}{\frac{1}{2}}=3\)

khi \(x=3+2\sqrt{2}\Rightarrow\sqrt{x}=\sqrt{2}+1\Rightarrow A=\frac{\sqrt{2}+1+1}{\sqrt{2}+1}=\sqrt{2}\)

Answer 2

[Hoàng Như Quỳnh]

\(a,ĐKXĐ:A=x\ge0;x\ne1\)

\(b,A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\)

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

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

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

\(A=\frac{\sqrt{x}+1}{\sqrt{x}}< =>ĐPCM\)

c,thay \(x=\frac{1}{4}\)vào A

\(c,A=\frac{\sqrt{\frac{1}{4}}+1}{\sqrt{\frac{1}{4}}}\)

\(A=\frac{\frac{1}{2}+1}{\frac{1}{2}}\)

\(A=3\)

\(x=3+2\sqrt{2}\)

\(x=\sqrt{2}^2+2\sqrt{2}+1\)

\(x=\left(\sqrt{2}+1\right)^2\)thay x vào A

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

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

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

\(A=\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\)