\(A=x-\sqrt{x}\)

\(+,0< x< 1\Rightarrow\sqrt{x}>x\Rightarrow x-\sqrt{x}< 0\Rightarrow A< 0\Rightarrow A< \left|A\right|\)

\(+,x\ge1\Rightarrow x\ge1\Rightarrow x\ge\sqrt{x}\Rightarrow x-\sqrt{x}\ge0\Rightarrow A\ge0\Rightarrow A=\left|A\right|\)

\(b,A=2\Leftrightarrow x-\sqrt{x}=2\Leftrightarrow x-\sqrt{x}+\frac{1}{4}=2+\frac{1}{4}=\frac{9}{4}\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2=\left(\pm\frac{3}{2}\right)^2\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=-1\left(loại\right)\end{matrix}\right.\Leftrightarrow x=4\) \(c,A=x-\sqrt{x}\Rightarrow A=x-\sqrt{x}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge0-\frac{1}{4}=\frac{-1}{4}\Rightarrow A_{min}=\frac{-1}{4}.\text{Dâu "=" xay ra khi:}\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)

ĐKXĐ: x>=0

a: P=1/2

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}+5}=\dfrac{1}{2}\)

=>\(2\sqrt{x}+4=\sqrt{x}+5\)

=>\(\sqrt{x}=1\)

=>x=1(nhận)

b: \(P^2-P=P\left(P-1\right)\)

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+5}\cdot\dfrac{\sqrt{x}+2-\sqrt{x}-5}{\sqrt{x}+5}\)

\(=\dfrac{-3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+5\right)^2}< 0\)

=>\(P^2< P\)

c: Để P nguyên thì \(\sqrt{x}+2⋮\sqrt{x}+5\)

=>\(\sqrt{x}+5-3⋮\sqrt{x}+5\)

=>\(\sqrt{x}+5\inƯ\left(-3\right)\)

=>\(\sqrt{x}+5\in\left\{1;-1;3;-3\right\}\)

=>\(\sqrt{x}\in\left\{-4;-6;-2;-8\right\}\)

=>\(x\in\varnothing\)

\(A=x-\sqrt{x}\)   \(\left(ĐKXĐ:x\ge0\right)\)

\(A=x-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\) 

\(A=\left(x-\frac{1}{2}\right)^2\) \(-\frac{1}{4}\) 

Có \(\left(x-\frac{1}{2^2}\right)\ge0\forall x\ge0\) 

     \(\left(x-\frac{1}{2}\right)^2\) -    1/4  >= \(\frac{-1}{4}\)mọi x>=0

   Dấu = sảy ra \(\Leftrightarrow\) x- \(\frac{1}{2}\) = 0

                        \(\Leftrightarrow\) x = 1 / 2  (  t/m  ) 

 vậy A đạt GTNN là -1/4 tại x = 1/2

Tớ nhầm nhé \(x\) từ dòng thứ 3 xuống pahir thay =\(\sqrt{x}\)

a) ĐKXĐ: \(x>0\)

\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1\)

\(=x+\sqrt{x}-2\sqrt{x}-1+1=x-\sqrt{x}\)

\(A=x-\sqrt{x}=2\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)(do \(\sqrt{x}+1\ge1>0\))

b) \(A=x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)>0\)(do \(x>1\))

\(\Leftrightarrow A=x-\sqrt{x}=\left|A\right|\)

c) \(A=x-\sqrt{x}=\left(x-\sqrt{x}+\dfrac{1}{4}\right)-\dfrac{1}{4}\)

\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(minA=-\dfrac{1}{4}\Leftrightarrow\sqrt[]{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)

\(a,A=\dfrac{x\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1\left(x>0\right)\\ A=\dfrac{x\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-2\sqrt{x}-1+1\\ A=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\\ A=2\Leftrightarrow x-\sqrt{x}-2=0\\ \Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}=2\left(\sqrt{x}>0\right)\\ \Leftrightarrow x=4\left(tm\right)\)

\(b,x>1\Leftrightarrow\sqrt{x}-1>0\\ \Leftrightarrow\left|A\right|=\left|x-\sqrt{x}\right|=\left|\sqrt{x}\left(\sqrt{x}-1\right)\right|=\sqrt{x}\left(\sqrt{x}-1\right)=A\left(\sqrt{x}>0\right)\)

\(c,A=x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\\ A_{min}=-\dfrac{1}{4}\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)

a: \(A=\sqrt{x}+\dfrac{\sqrt{x}\left(1+2\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\sqrt{x}+\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)

Khi x=4 thì \(A=2+\dfrac{2\cdot2+1}{2+1}=2+\dfrac{5}{3}=\dfrac{11}{3}\)

b: Khi x=(2-căn 3)^2 thì \(A=2-\sqrt{3}+\dfrac{2\left(2-\sqrt{3}\right)+1}{2-\sqrt{3}+1}\)

\(=2-\sqrt{3}+\dfrac{4-2\sqrt{3}+1}{3-\sqrt{3}}\)

\(=2-\sqrt{3}+\dfrac{5-2\sqrt{3}}{3-\sqrt{3}}\)

\(=\dfrac{\left(2-\sqrt{3}\right)\left(3-\sqrt{3}\right)+5-2\sqrt{3}}{3-\sqrt{3}}\)

\(=\dfrac{6-2\sqrt{3}-3\sqrt{3}+3+5-2\sqrt{3}}{3-\sqrt{3}}\)

\(=\dfrac{14-7\sqrt{3}}{3-\sqrt{3}}\)

d: A=2

=>\(\dfrac{x+\sqrt{x}+2\sqrt{x}+1}{\sqrt{x}+1}=2\)

=>\(x+3\sqrt{x}+1=2\left(\sqrt{x}+1\right)=2\sqrt{x}+2\)

=>\(x+\sqrt{x}-1=0\)

=>\(\left[{}\begin{matrix}\sqrt{x}=\dfrac{-1+\sqrt{5}}{2}\left(nhận\right)\\\sqrt{x}=\dfrac{-1-\sqrt{5}}{2}\left(loại\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{6-2\sqrt{5}}{4}=\dfrac{3-\sqrt{5}}{2}\)

a, ĐKXĐ : \(\left\{{}\begin{matrix}\dfrac{3x-5}{x-1}\ge0\\x-1\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-5\ge0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}3x-5\le0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{3}\\x>1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{3}\\x< 1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{3}\\x< 1\end{matrix}\right.\)

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b, Ta có : \(A=\sqrt{\dfrac{3x-5}{x-1}}=3\)

\(\Leftrightarrow3x-5=9x-9\)

\(\Leftrightarrow x=\dfrac{2}{3}\left(TM\right)\)

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