\(a,P=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\left(x\ge0;x\ne1\right)\\ P=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(x+16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\\ P=\dfrac{x+16}{\sqrt{x}+3}\\ b,P=4\Leftrightarrow\dfrac{x+16}{\sqrt{x}+3}=4\\ \Leftrightarrow x+16=4\sqrt{x}+12\\ \Leftrightarrow x-4\sqrt{x}+4=0\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\\ \Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)

\(c,P=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\\ P=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{25}{\sqrt{x}+3}}-6=2\cdot5-6=4\\ P_{min}=4\Leftrightarrow\left(\sqrt{x}+3\right)^2=25\Leftrightarrow\sqrt{x}+3=5\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow x=4\left(tm\right)\)

\(d,x=3-2\sqrt{2}\Leftrightarrow\sqrt{x}=\sqrt{2}-1\\ \Leftrightarrow P=\dfrac{3-2\sqrt{2}+16}{\sqrt{2}-1+3}=\dfrac{19-2\sqrt{2}}{\sqrt{2}+2}\\ P=\dfrac{\left(19-2\sqrt{2}\right)\left(2-\sqrt{2}\right)}{2}=\dfrac{42-23\sqrt{2}}{2}\)

a: \(=x\sqrt{2}-\sqrt{\left(x\sqrt{2}+1\right)^2}=x\sqrt{2}-\left|x\sqrt{2}+1\right|\)

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

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

\(\Leftrightarrow2x\sqrt{2}=-4\)

hay \(x=-\sqrt{2}\)

a) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)

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

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

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

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

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

b) Ta có: \(x=3-2\sqrt{2}\)

\(=2-2\cdot\sqrt{2}\cdot1+1\)

\(=\left(\sqrt{2}-1\right)^2\)

Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(P=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\), ta được: 

\(P=\dfrac{\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)^2}}\)

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

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

\(=\dfrac{1}{\sqrt{2}-1}\)

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

Vậy: Khi \(x=3-2\sqrt{2}\) thì \(P=\sqrt{2}+1\)

Lời giải:

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

b. $B=3\Leftrightarrow \frac{\sqrt{x}}{1-\sqrt{x}}=3$

$\Rightarrow \sqrt{x}=3(1-\sqrt{x})$

$\Leftrightarrow 4\sqrt{x}=3\Leftrightarrow x=\frac{9}{16}$ (tm) 

c.

Khi $x=3-2\sqrt{2}=(\sqrt{2}-1)^2\Rightarrow \sqrt{x}=\sqrt{2}-1$

Khi đó:

$B=\frac{\sqrt{x}}{1-\sqrt{x}}=\frac{\sqrt{2}-1}{1-(\sqrt{2}-1)}=\frac{\sqrt{2}-1}{2-\sqrt{2}}$

\(a,B=\dfrac{-\sqrt{x}-3+\sqrt{x}-3+x+4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\left(x\ge0;x\ne9\right)\\ B=\dfrac{x-2}{x-9}=\dfrac{x-9+7}{x-9}=1+\dfrac{7}{x-9}\in Z\\ \Leftrightarrow x-9\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{2;8;11;16\right\}\)

Vậy giá trị x thỏa đề là \(x=2\)

\(a,E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\dfrac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\left(x>0;x\ne1\right)\\ E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\dfrac{x}{\sqrt{x}-1}\\ b,E>1\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\\ \Leftrightarrow\sqrt{x}-1>0\left[x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\right]\\ \Leftrightarrow x>1\left(tm\right)\)

\(c,E=\dfrac{x}{\sqrt{x}-1}=\dfrac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}\\ E=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\ge2\sqrt{\dfrac{\sqrt{x}-1}{\sqrt{x}-1}}+2=2+2=4\\ E_{min}=4\Leftrightarrow\sqrt{x}-1=1\Leftrightarrow x=4\)