Question

cho  p=\(\left(\dfrac{\sqrt{x}}{\sqrt{x-2}}+\dfrac{1}{x-4}\right)\) :\(\dfrac{\sqrt{x}+1}{x-4}\)
a, đkxđ
b, rút gọn 
c, tính p khi x=3-\(\sqrt{8}\)

Answer 1

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne2\\x\ne4\\x\ge0\end{matrix}\right.\)

 

Answer 2

a, ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x-2>0\\x-4\ne0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x>2\\x\ne4\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>2\\x\ne4\end{matrix}\right.\)

mik thấy đề sai sai

Answer 3

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

b: Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{1}{x-4}\right):\dfrac{\sqrt{x}+1}{x-4}\)

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

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