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a)\(ĐKXĐ\Leftrightarrow\begin{cases}\sqrt{x}\ge0\\\sqrt{x}-1\ne0\end{cases}\Leftrightarrow\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(A=\frac{\sqrt{x}\cdot\left(\sqrt{x}+2\right)+1\cdot\left(\sqrt{x}-1\right)-3\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}+\sqrt{x}-1-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
b)\(S=A\cdot B\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\frac{\sqrt{x}+3}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+3}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}+2+1}{\sqrt{x}+2}\)
\(=1+\frac{1}{\sqrt{x}+2}\)
Để S đạt GTLN thì \(\frac{1}{\sqrt{x}+2}\) đạt GTLN
\(\frac{1}{\sqrt{x}+2}\) đạt GTLN \(\Leftrightarrow\sqrt{x}+2\) đạt GTNN
GTNN \(\sqrt{x}+2\) là 2 \(\Leftrightarrow x=0\)
Vậy GTLN của S là \(\frac{3}{2}\Leftrightarrow x=0\)
a/ \(A=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}-\frac{3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\) \(\left(ĐK:x\ge0;x\ne1\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}-1-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}+\sqrt{x}-1-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
a) ĐKXĐ : \(x\ne4\)
\(A=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-5-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)\(=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
b) Biểu thức B xác định \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne16\end{matrix}\right.\)
+ \(x^2-9=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=-3\left(KTM\right)\end{matrix}\right.\)
+ Với x = 3 ta có :
\(B=\frac{2\sqrt{3}+5}{\sqrt{3}-4}\)\(=\frac{-\left(\sqrt{3}-4\right)\left(\sqrt{3}+2\right)}{\sqrt{3}-4}=-2-\sqrt{3}\)
c) \(A\cdot B=\frac{\sqrt{x}-4}{\sqrt{x}-2}\cdot\frac{2\sqrt{x}+5}{\sqrt{x}-4}=\frac{2\sqrt{x}+5}{\sqrt{x}-2}=2+\frac{9}{\sqrt{x}-2}\)
\(\Rightarrow A\cdot B\) là số tự nhiên \(\Rightarrow\left\{{}\begin{matrix}9⋮\sqrt{x}-2\\\frac{9}{\sqrt{x}-2}\ge-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-2\in\left\{9;3;1;-1\right\}\\\frac{9}{\sqrt{x}-2}\ge-2\end{matrix}\right.\)
\(\Rightarrow x\in\left\{121;25;9;\right\}\)