mik cần gấp ạ^^

a: \(=6+2\sqrt{11}-4+\sqrt{11}=2+3\sqrt{11}\)

b: \(=\dfrac{3x+9\sqrt{x}-2x+4\sqrt{x}}{\left(\sqrt{x}+3\right)\left(x-2\sqrt{x}\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}=\dfrac{\sqrt{x}+3}{x-2\sqrt{x}}\)

a) \(P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

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

\(\Leftrightarrow-\sqrt{x}-1=\sqrt{x}-1\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)

c) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\)

\(\Rightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Kết hợp đk:

\(\Rightarrow x\in\left\{0\right\}\)

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

\(a,P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,P=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\\ \Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\\ c,P\in Z\Leftrightarrow\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}\in Z\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\in Z\\ \Leftrightarrow2⋮\sqrt{x}+1\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}+1\in\left\{1;2\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}\in\left\{0;1\right\}\\ \Leftrightarrow x\in\left\{0;1\right\}\)

\(d,P=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)

Có \(\dfrac{2}{\sqrt{x}+1}>0\left(2>0;\sqrt{x}+1>0\right)\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}< 1\Leftrightarrow P< 1\)

\(e,P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)

Có \(\sqrt{x}+1\ge1\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\le2\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\ge1-2=-1\)

\(P_{min}=-1\Leftrightarrow x=0\)

a) Thay x=25 vào B ta có:

\(B=\dfrac{\sqrt{25}+2}{\sqrt{25}-2}=\dfrac{7}{3}\)

b) \(A=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{2\sqrt{x}-1}{x-5\sqrt{x}+6}\)

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

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

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

\(A=\dfrac{2\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(A=\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

c) Ta có: \(A>B\) Khi:

\(\dfrac{2}{\sqrt{x}-2}>\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)

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

\(\Leftrightarrow\dfrac{2-\sqrt{x}-2}{\sqrt{x}-2}>0\)

\(\Leftrightarrow\dfrac{-\sqrt{x}}{\sqrt{x}-2}>0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}-\sqrt{x}< 0\\\sqrt{x}-2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}-\sqrt{x}>0\\\sqrt{x}-2>0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x>4\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow0< x< 4\) 

\(B=\left(\dfrac{3\sqrt{x}+6}{x-4}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{x-9}{\sqrt{x}-3}\left(x\ge0;x\ne4;x\ne9\right)\)

\(=\left[\dfrac{3\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\cdot\dfrac{\sqrt{x}-3}{x-9}\)

\(=\dfrac{3\sqrt{x}+6+x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+5\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{1}{\sqrt{x}+3}\)

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

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

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

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

#\(Toru\)

Lời giải:

$A=\frac{10\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+4)}-\frac{(2\sqrt{x}-3)(\sqrt{x}-1)}{(\sqrt{x}+4)(\sqrt{x}-1)}-\frac{(\sqrt{x}+1)(\sqrt{x}+4)}{(\sqrt{x}-1)(\sqrt{x}+4)}$

$=\frac{10\sqrt{x}-(2\sqrt{x}-3)(\sqrt{x}-1)-(\sqrt{x}+1)(\sqrt{x}+4)}{(\sqrt{x}+4)(\sqrt{x}-1)}$

$=\frac{-3x+10\sqrt{x}-7}{(\sqrt{x}+4)(\sqrt{x}-1)}$

$=\frac{-(\sqrt{x}-1)(3\sqrt{x}-7)}{(\sqrt{x}+4)(\sqrt{x}-1)}=\frac{7-3\sqrt{x}}{\sqrt{x}+4}$

a) \(B=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

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

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

\(=\dfrac{-3}{\sqrt{x}+3}:\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=-\dfrac{3}{\sqrt{x}+3}.\dfrac{\sqrt{x}+3}{-\left(\sqrt{x}-2\right)}\)

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

b) \(\sqrt{x}=\sqrt{7-4\sqrt{3}}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

Thế vào B \(\Rightarrow B=\dfrac{3}{2-\sqrt{3}-2}=\dfrac{3}{-\sqrt{3}}=-\sqrt{3}\)

a) Ta có: \(B=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\dfrac{x-3\sqrt{x}-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{9-x+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{-3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{-x+4\sqrt{x}-4}\)

\(=\dfrac{-3\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}=\dfrac{3}{\sqrt{x}-2}\)

b) Thay \(x=7-4\sqrt{3}\) vào B, ta được:

\(B=\dfrac{3}{2-\sqrt{3}-2}=-\sqrt{3}\)

\(a)E=\left(\dfrac{x-2\sqrt{x}}{x-4}-1\right):\left(\dfrac{4-x}{x-\sqrt{x}-6}+\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\right)\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-1\right):\left(\dfrac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+2}\right):\dfrac{4-x+x-4-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}+2}:\dfrac{9-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{-2}{\sqrt{x}+2}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{9-x}\\ =\dfrac{-2\left(\sqrt{x}-3\right)}{9-x}=\dfrac{2\left(\sqrt{x}-3\right)}{x-9}\\ =\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{2}{\sqrt{x}+3}\)

\(b)\)E dương

\(\Leftrightarrow E>0\\ \Leftrightarrow\dfrac{2}{\sqrt{x}+3}>0\\ \Leftrightarrow\sqrt{x}+3>0\left(Vì.2>0\right)\\ \Leftrightarrow\sqrt{x}>-3\forall x\in R\\ \Rightarrow x\ge0\)

Kết hợp đk

\(x\ge0;x\ne4;x\ne9\) 

Vậy \(x\ge0;x\ne4;x\ne9\) thì E dương

a: Ta có: \(x=\sqrt{28-16\sqrt{3}}+2\sqrt{3}\)

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

=4

Thay x=4 vào B, ta được:

\(B=\dfrac{2-4}{2}=-1\)

\(A=\dfrac{x-4}{x}\cdot\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\right)\)

\(=\dfrac{x-4}{x}\cdot\dfrac{x-3\sqrt{x}+2-x-3\sqrt{x}-2}{x-4}=\dfrac{-6}{\sqrt{x}}\)