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a) ĐKXĐ: \(x>0;x\ne9\)
\(A=\left(\frac{1}{\sqrt{x}+3}+\frac{3}{x-9}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\frac{\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\frac{1}{\sqrt{x}+3}\)
a) \(P=\left[\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-\left(3x+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\left[\frac{\left(2\sqrt{x}-2\right)-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\right]\left(ĐK:x\ge0;x\ne9\right)\)
\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{-3}{\sqrt{x}+3}\)
a) \(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{4\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{4\sqrt{x}-12}{x-9}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{x+3\sqrt{x}}{x-9}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}\)
\(=\frac{x-25}{x-9}\)
b) \(P=\frac{A}{B}=\frac{\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}\)
\(=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
\(\sqrt{P}< \frac{1}{3}\Rightarrow\sqrt{\frac{\sqrt{x}-5}{\sqrt{x}+3}}< \frac{1}{3}\)
\(\Rightarrow\frac{\sqrt{x}-5}{\sqrt{x}+3}< \frac{1}{9}\Leftrightarrow9\sqrt{x}-45< \sqrt{x}+3\)
\(\Leftrightarrow8\sqrt{x}< 48\Leftrightarrow\sqrt{x}< 6\Rightarrow0\le x< 36\)
\(a,\)\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)\(=\frac{2x+3\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(b,P=\frac{A}{B}=\frac{2x+3\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{\sqrt{x}+5}{\sqrt{x}-3}\)
\(=\frac{2x+3\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\left(\sqrt{x}-3\right)}{\sqrt{x}+5}=\frac{2x+3\sqrt{x}-1}{\sqrt{x}+5}\)
Để \(\sqrt{P}< \frac{1}{3}\Rightarrow\frac{2x+3\sqrt{x}-1}{\sqrt{x}+5}< \frac{1}{3}\)
\(\Rightarrow\frac{2x+3\sqrt{x}-1}{\sqrt{x}+5}-\frac{1}{3}< 0\)
\(\Rightarrow\frac{3\left(2x+3\sqrt{x}-1\right)-\sqrt{x}-5}{3\left(\sqrt{x}+5\right)}< 0\)
\(\Rightarrow6x+9\sqrt{x}-3-\sqrt{x}-5< 0\)( do \(3\left(\sqrt{x}+5\right)>0\))
\(\Rightarrow6x-8\sqrt{x}-8< 0\Rightarrow3x-4\sqrt{x}-4< 0\)
\(\Rightarrow3x-6\sqrt{x}+2\sqrt{x}-4< 0\)
\(\Rightarrow3\sqrt{x}\left(\sqrt{x}-2\right)+2\left(\sqrt{x}-2\right)< 0\)
\(\Rightarrow\left(\sqrt{x}-2\right)\left(3\sqrt{x}+2\right)< 0\)
Vì \(3\sqrt{x}+2>0\Rightarrow\sqrt{x}-2< 0\)
\(\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
Vậy để \(\sqrt{P}< \frac{1}{3}\)thì \(0\le x< 4\)
a) đk: \(x\ge0\)
Ta có: \(P< -\frac{1}{3}\)
\(\Leftrightarrow\frac{-3}{\sqrt{x}+3}+\frac{1}{3}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-6}{\sqrt{x}+3}< 0\)
Nhận thấy \(\sqrt{x}-6< \sqrt{x}+3\)
\(\Rightarrow\hept{\begin{cases}\sqrt{x}-6< 0\\\sqrt{x}+3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}< 6\\\sqrt{x}>-3\end{cases}}\Leftrightarrow0\le x< 36\)
b) Ta có: \(Q=5+P=5-\frac{3}{\sqrt{x}+3}\)
Mà \(\sqrt{x}+3\ge3\left(\forall x\ge0\right)\Leftrightarrow\frac{3}{\sqrt{x}+3}\le1\left(\forall x\ge0\right)\)
\(\Leftrightarrow5-\frac{3}{\sqrt{x}+3}\ge4\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(x=0\)
Vậy Min(Q) = 4 khi x = 0