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

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

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

a) Rút gọn P

ĐKXĐ: \(x\ge0;x\ne1\)

\(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)\(-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)\(-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)\(=\dfrac{\left(-5\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

b) Tìm GTLN

\(P=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=\dfrac{17-5\left(\sqrt{x}+3\right)}{\sqrt{x}+3}=\dfrac{17}{\sqrt{x}+3}-5\)

Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\Rightarrow P=\dfrac{17}{\sqrt{x}+3}-5\le\dfrac{17}{3}-5=\dfrac{2}{3}\)

Dấu "=" xảy ra khi \(x=0\)

Vậy \(P_{max}=\dfrac{2}{3}\) khi \(x=0\)

a) \(P=\dfrac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

\(=\dfrac{3x+3\sqrt{x}-9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(=\dfrac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)

b) \(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+2}{\sqrt{x}+2}=3+\dfrac{2}{\sqrt{x}+2}\)

Để \(P\in Z\Rightarrow2⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2=2\left(\sqrt{x}+2\ge2\right)\)

\(\Rightarrow x=0\)

c) Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\Rightarrow3+\dfrac{2}{\sqrt{x}+2}\le4\)

\(\Rightarrow P_{max}=4\) khi \(x=0\)

Lời giải: 
ĐKXĐ: $x\geq 0; x\neq 1$
a.

\(A=\left[\frac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\right].\frac{2}{\sqrt{x}-1}\)

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

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

b.

Ta thấy với $x\geq 0 ; x\neq 1$ thì $x+\sqrt{x}+1\geq 1$

$\Rightarrow A=\frac{2}{x+\sqrt{x}+1}\leq 2$

Vậy $A$ đạt max bằng $2$ khi $x=0$

\(a,A=\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{x-2-x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

a) \(A=\dfrac{3x}{x\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{1}{1+\sqrt{x}}\)

\(=\dfrac{3x}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{1}{1+\sqrt{x}}\)

\(=\dfrac{3x-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

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

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

\(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}}\right).\dfrac{\left(1-x\right)^2}{2}\) (ĐK:\(x>0;x\ne1\))

\(=\left[\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)}\right].\dfrac{\left(x-1\right)^2}{2}\)

\(=\left[\dfrac{\left(\sqrt{x}-2\right)\sqrt{x}}{\left(x-1\right)\sqrt{x}}-\dfrac{x-1}{\sqrt{x}\left(x-1\right)}\right].\dfrac{\left(x-1\right)^2}{2}\)

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

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

Sai đề ko em?

a) Ta có: \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}}\right)\cdot\dfrac{\left(1-x\right)^2}{2}\)

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

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

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

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

=x−2√x−x+1√x(x−1).(x−1)22=−2√x+1√x(x−1).(x−1)22=x−2x−x+1x(x−1).(x−1)22=−2x+1x(x−1).(x−1)22

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

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

c: Ta có: \(x+\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ

\(\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}>0\forall x\)

\(a,P\) xác định \(\Leftrightarrow\left[{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)

\(b,P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\\ =\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\\ =\dfrac{1}{\sqrt{x}}.\dfrac{\sqrt{x}-2}{3}\\ =\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

\(c,P=\dfrac{1}{4}\Leftrightarrow\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{4\left(\sqrt{x}-2\right)-3\sqrt{x}}{12\sqrt{x}}=0\\ \Leftrightarrow4\sqrt{x}-8-3\sqrt{x}=0\\ \Leftrightarrow\sqrt{x}=8\\ \Leftrightarrow x=64\left(tmdk\right)\)

Vậy \(x=64\) thì \(P=\dfrac{1}{4}\)

Lời giải:

ĐK: $x\geq 0; x\neq 1$

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

\(=\frac{(\sqrt{x}-2)(\sqrt{x}-1)}{2(\sqrt{x}+1)}-\frac{\sqrt{x}+2}{2}=\frac{(\sqrt{x}-2)(\sqrt{x}-1)-(\sqrt{x}+2)(\sqrt{x}+1)}{2(\sqrt{x}+1)}=\frac{-6\sqrt{x}}{2(\sqrt{x}+2)}=\frac{-3\sqrt{x}}{\sqrt{x}+2}\)

Vì $x\geq 0$ nên $3\sqrt{x}\geq 0; \sqrt{x}+2>0$

$\Rightarrow \frac{3\sqrt{x}}{\sqrt{x}+2}\geq 0$

$\Rightarrow A\leq 0$ hay $A_{\max}=0$ khi $x=0$