a) đkxđ: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}+1\ne0;x-\sqrt{x}+1\ne0\end{matrix}\right.\)
b)Rgọn
A=\(x-\frac{2x-2\sqrt{x}}{\sqrt{x}-1}+\frac{x\sqrt{x}+1}{x-\sqrt{x}+1}+1\)
<=> \(x-\frac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1\)
<=>\(x-2\sqrt{x}+\sqrt{x}+1+1\)
<=> \(x-\sqrt{x}+2\)
Đến đây bạn tự lm nhé.
Chúc bạn học tốt
Nhớ tích cho mình nhé
a) ĐKXĐ:\(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b)\(A=x-\frac{2x-2\sqrt{x}}{\sqrt{x}-1}+\frac{x\sqrt{x}+1}{x-\sqrt{x}+1}+1\)
\(=x-\frac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\frac{\left(\sqrt{x}\right)^3+1}{x-\sqrt{x}+1}+1\)
\(=x-2\sqrt{x}+\sqrt{x}+1+1\)
\(=x-\sqrt{x}+2\)
Ta có: \(x-\sqrt{x}+2=x-2.\frac{1}{2}\sqrt{x}+\frac{1}{4}-\frac{1}{4}+2\)
\(=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}\)
Vì \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0,\forall x\)
\(\Rightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)
\(\Rightarrow Min_A=\frac{7}{4}\) khi \(\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)
