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Đk: \(x\ge0\)
a) Ta có: x = 16 => A = \(\frac{\sqrt{16}+5}{\sqrt{16}+2}=\frac{4+5}{4+2}=\frac{9}{6}=\frac{3}{2}\)
\(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\)=> \(\sqrt{x}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)
=> A = \(\frac{\sqrt{2}-1+5}{\sqrt{2}-1+2}=\frac{\sqrt{2}+4}{\sqrt{2}+2}=\frac{\sqrt{2}\left(2\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{\sqrt{2}\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\frac{4-\sqrt{2}-1}{2-1}=3-\sqrt{2}\)
b) A = 2 <=> \(\frac{\sqrt{x}+5}{\sqrt{x}+2}=2\) <=> \(\sqrt{x}+5=2\sqrt{x}+4\) <=> \(\sqrt{x}=1\) <=> x = 1 (tm)
\(A=\sqrt{x}+1\) <=> \(\frac{\sqrt{x}+5}{\sqrt{x}+2}=\sqrt{x}+1\) <=> \(\sqrt{x}+5=\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\)
<=> \(\sqrt{x}+5=x+3\sqrt{x}+2\) <=> \(x+2\sqrt{x}-3=0\)<=> \(x+3\sqrt{x}-\sqrt{x}-3=0\)
<=> \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)=0\) <=> \(\sqrt{x}-1=0\)(vì \(\sqrt{x}+3>0\))
<=> \(x=1\)(tm)
c) Ta có: \(A=\frac{\sqrt{x}+5}{\sqrt{x}+2}=\frac{\sqrt{x}+2+3}{\sqrt{x}+2}=1+\frac{3}{\sqrt{x}+2}\)
Do \(\sqrt{x}+2\ge\) => \(\frac{3}{\sqrt{x}+2}\le\frac{3}{2}\) => \(1+\frac{3}{\sqrt{x}+2}\le1+\frac{3}{2}=\frac{5}{2}\) => A \(\le\)5/2
Dấu "=" xảy ra<=> x = 0
Vậy MaxA = 5/2 <=> x = 0
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\right)\)
\(\Leftrightarrow P=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{x\left(\sqrt{x}+1\right)}=\frac{x}{\sqrt{x}-1}\)
b. ta có \(x=\frac{8-4\sqrt{3}}{2-\sqrt{3}}=4\)
vậy \(P=\frac{4}{\sqrt{4}-1}=4\)
c.\(P=\frac{x}{\sqrt{x}-1}=\sqrt{x}-1+\frac{1}{\sqrt{x}-1}+2\ge2+2=4\)
vậy \(\sqrt{P}\ge2\)
Điều kiện có nghĩa
a/ \(\hept{\begin{cases}x+2\ge0\\x-5\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge-2\\x\ne5\end{cases}}\)
b/ \(\hept{\begin{cases}2x-1\ge0\\x+3\ne0\end{cases}}\)
\(\hept{\begin{cases}x\ge\frac{1}{2}\\x\ne-3\end{cases}}\)
c/ \(\left(x-3\right)\left(x+2\right)\ge0\)
\(\Leftrightarrow\orbr{\begin{cases}x\le-2\\x\ge3\end{cases}}\)
d/ \(\hept{\begin{cases}2x-1\ge0\\-x\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge\frac{1}{2}\\x\le0\end{cases}}\)
Không tồn tại x để nó có nghĩa.
e/ \(\hept{\begin{cases}-3x\ge0\\x+2>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\le0\\x>-2\end{cases}}\)