Question

Với x\(\ge\)0,x khác 9 cho biểu thức P = \(\frac{x-4\sqrt{x}+4}{\sqrt{x}-3}\)tìm x để P \(\le\)0

Answer 1

Ta có:\(P=\frac{x-4\sqrt{x}+4}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-3}\)

Để \(P\le0\) mà \(\left(\sqrt{x}-2\right)^2\ge0\) nên \(\sqrt{x}-3< 0\)

\(\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\)

Để \(P\le0\) thì \(0\le x< 9\)

Answer 2

\(P=\frac{x-4\sqrt{x}+4}{\sqrt{x}-3}\le0\\ P=\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-3}\le0\)

Vì \(\left(\sqrt{x}-2\right)^2\ge0\forall x\ne9,x\ge0\)nên :

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}\left(\sqrt{x}-2\right)^2\ne0\\\sqrt{x}-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\left(\sqrt{x}-2\right)^2=0\\\sqrt{x}-3\ne0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ne4\\x< 9\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\\x\ne9\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x< 9,x\ne4\\x=4\end{matrix}\right.\\ Vậy..\)

Kết hợp đkxđ \(0\le x< 9,x\ne4hoặcx=4\)

vậy ...