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ĐKXĐ: x \(\ge\)0; x \(\ne\)1 ; x \(\ne\)4
a) P = \(\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\)
P = \(\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\sqrt{x}+1}:\frac{\sqrt{x}\left(1-\sqrt{x}\right)-\sqrt{x}+4}{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}\)
P = \(\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}\cdot\frac{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-x-\sqrt{x}+4}\)
P = \(\frac{\left(1-\sqrt{x}\right)\left(\sqrt{x}-2\right)}{4-x}\)
P = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
b) P < 0 <=> \(\frac{\sqrt{x}-1}{\sqrt{x}+2}< 0\)
Do \(\sqrt{x}+2>0\) => \(\sqrt{x}-1< 0\) => \(\sqrt{x}< 1\) => \(x< 1\)
kết hợp với đk => S = {x| \(0\le x< 1\)}
c) P = \(\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\ge-\frac{1}{2}\)
Do \(\sqrt{x}+2\ge2\) => \(-\frac{3}{\sqrt{x}+2}\ge-\frac{3}{2}\) => \(1-\frac{3}{\sqrt{x}+2}\ge-\frac{1}{2}\)
Dấu "=" xảy ra <=> x = 0
Vậy MinP = -1/2 khi x = 0
a: Ta có: \(P=\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{\sqrt{x}-1}{\sqrt{x}-x}+\dfrac{\sqrt{x}+3}{x+5\sqrt{x}+6}\)
\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-2-\sqrt{x}-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
a) Ta có:
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right)\div\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)
\(P=\frac{x-1}{\left(\sqrt{x}-1\right)\sqrt{x}}\div\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\sqrt{x}}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{x-1}{\sqrt{x}}\)
b) Ta có: \(P>0\)
\(\Leftrightarrow\frac{x-1}{\sqrt{x}}>0\)
\(\Leftrightarrow\frac{\left(x-1\right)\sqrt{x}}{x}>0\)
\(\Rightarrow\left(x-1\right)\sqrt{x}>0\)
\(\Rightarrow\hept{\begin{cases}x-1>0\\\sqrt{x}>0\end{cases}}\Rightarrow x>1\)
Vậy khi \(x>1\Leftrightarrow P>0\)
c) Ta có: \(P=6\)
\(\Leftrightarrow\frac{x-1}{\sqrt{x}}=6\)
\(\Leftrightarrow x-1=6\sqrt{x}\)
\(\Leftrightarrow\left(x-1\right)^2=36x\)
\(\Leftrightarrow x^2-38x+1=0\)
\(\Leftrightarrow\left(x^2-38x+361\right)-360=0\)
\(\Leftrightarrow\left(x-19\right)^2-\left(6\sqrt{10}\right)^2=0\)
\(\Leftrightarrow\left(x-19-6\sqrt{10}\right)\left(x-19+6\sqrt{10}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-19-6\sqrt{10}=0\\x-19+6\sqrt{10}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=19+6\sqrt{10}\\x=19-6\sqrt{10}\end{cases}}\)
Chi biet phan 5 thoi @
Vi 3a=5b=12suy ra a=4 ;b=2,4 ta co p=a.b suy ra p=4×2.4=9.6 suy ra p>[=9.6 gtln=9.6