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1) ĐK \(x\ne\pm9\)
\(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3x+9}{x-9}=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\frac{2\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\frac{3x+9}{x-9}\)
\(=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3}{\sqrt{x}+3}\)
2) ?
3) Ta có
\(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+3\ge3\)
\(\Rightarrow A=\frac{3}{\sqrt{x}+3}\le1\)
Dấu "=" xảy ra khi x = 0
\(P=\frac{\sqrt{x}-1}{\sqrt{x}}-9\sqrt{x}=1-\left(\frac{1}{\sqrt{x}}+9\sqrt{x}\right)\)
\(\frac{1}{\sqrt{x}}+9\sqrt{x}\ge2\sqrt{\frac{1}{\sqrt{x}}\cdot9\sqrt{x}}=6\)
\(\Rightarrow P\le1-6=-5\)
Dấu "=" xảy ra khi \(\frac{1}{\sqrt{x}}=9\sqrt{x}\Leftrightarrow x=\frac{1}{9}\)
Vậy MaxP =-5 đạt được khi \(x=\frac{1}{9}\)
a/ Bạn tự giải
b/ \(B=\frac{x-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(B=\frac{x-7+\sqrt{x}-3-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{x-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-1}\)
c/ \(P=AB=\left(\frac{\sqrt{x}-1}{\sqrt{x}+2}\right)\left(\frac{\sqrt{x}+3}{\sqrt{x}-1}\right)=\frac{\sqrt{x}+3}{\sqrt{x}+2}=1+\frac{1}{\sqrt{x}+2}\)
Do \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\frac{1}{\sqrt{x}+2}\le\frac{1}{2}\)
\(\Rightarrow P\le1+\frac{1}{2}=\frac{3}{2}\Rightarrow P_{max}=\frac{3}{2}\) khi \(x=0\)
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/ \(A=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}-\frac{3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\) \(\left(ĐK:x\ge0;x\ne1\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}-1-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}+\sqrt{x}-1-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
P=\(\frac{\sqrt{x}-1-9x}{\sqrt{x}}=\frac{-5\sqrt{x}-\left(9x-6\sqrt{x}+1\right)}{\sqrt{x}}=-5-\frac{\left(3\sqrt{x}-1\right)^2}{\sqrt{x}}\le-5\)
Dấu "=" xảy ra\(\Leftrightarrow3\sqrt{x}-1=0\Leftrightarrow x=\frac{1}{9}\)
Vậy: Pmax = -5 \(\Leftrightarrow x=\frac{1}{9}\)
Ta có
\(x^2\ge0\) với mọi x
\(\Rightarrow x^2+1\ge1\)
\(\Rightarrow\sqrt{x^2+1}\ge1\)
\(\Rightarrow\frac{1}{\sqrt{x^2+1}}\le1\)
\(\Rightarrow\frac{9}{\sqrt{x^2+1}}\le9\)
\(\Rightarrow1+\frac{9}{\sqrt{x^2+1}}\le10\)
Dấu " = " xảy ra khi x=0
Vậy MAXP=10 khi x=0
Để P đạt GTLN
\(\Rightarrow\sqrt{x^2+1}\) đạt GTNN
Ta thấy:\(x^2\ge0\)
\(\Rightarrow x^2+1\ge0+1=1\)
\(\Rightarrow\sqrt{x^2+1}\ge\sqrt{1}=1\)
Khi đó GTLN của P là \(1+\frac{9}{1}=1+9=10\) khi x=0
Vậy MaxP=10 khi x=0