BT1
a.A=\(\frac{2+\sqrt{x}}{x-9}\)
b Q=\(\frac{\sqrt{x}-1}{x}+\frac{2\sqrt{x}-1}{x+\sqrt{x}}\)
c.M=\(\frac{2+3\sqrt{x}}{x-5}\)
BT2
Cho P=\(\left(1+\frac{1}{\sqrt{x}-1}\right)-\frac{1}{x-\sqrt{x}}\)
a Tìm đkxd
b rút gọn
c tính p khi x=25
d Tính P khi \(x=\sqrt{5+2\sqrt{6}}\)
Câu 1: Điều kiện xác định
a/ \(\hept{\begin{cases}x\ge0\\x-9\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}}\)
b/ \(Q=\frac{\sqrt{x}-1}{x}+\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(\hept{\begin{cases}x>0\\\sqrt{x}+1\ne0\end{cases}\Rightarrow x>0}\)
c/ \(\hept{\begin{cases}x\ge0\\x-5\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\x\ne5\end{cases}}}\)
Câu 2:
a/ ĐKXĐ: \(\hept{\begin{cases}x>0\\\sqrt{x}-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x\ne1\end{cases}}}\)
b/ \(P=\left(1+\frac{1}{\sqrt{x}-1}\right)-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{x-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
c/ Thay x = 25 vào P ta được: \(P=\frac{\sqrt{25}+1}{\sqrt{25}}=\frac{6}{5}\)
d/ Ta có: \(P=\frac{\sqrt{5+2\sqrt{6}}+1}{\sqrt{5+2\sqrt{6}}}=\frac{\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+1}{\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}=\frac{\sqrt{3}+\sqrt{2}+1}{\sqrt{3}+\sqrt{2}}\)