a: \(Q=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

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

b: Khi x=4+2căn 3 thì \(Q=\dfrac{\sqrt{3}+1-2}{\sqrt{3}+1+2}=\dfrac{-3+2\sqrt{3}}{3}\)

c: Q=3

=>3căn x+6=căn x-2

=>2căn x=-8(loại)

d: Q>1/2

=>Q-1/2>0

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{1}{2}>0\)

=>2căn x-4-căn x-2>0

=>căn x>6

=>x>36

d: Q nguyên

=>căn x+2-4 chia hết cho căn x+2

=>căn x+2 thuộc Ư(-4)

=>căn x+2 thuộc {2;4}

=>x=0 hoặc x=4(nhận)

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

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

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

b) Q > 0

⇔ \(\dfrac{\sqrt{\text{x}}-2}{3\sqrt{\text{x}}}\) > 0

Do \(\text{3}\sqrt{\text{x}}>0\)   ∀x⩾0

⇒ \(\sqrt{\text{x}}-2>0\)

⇔ \(\sqrt{\text{x}}>2\)

⇔ x > 4

Vậy x > 4 thì Q > 0 

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

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

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

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

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

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

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

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

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

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

a) ĐKXĐ: \(x>0;x\ne4\)

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

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

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

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

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

b) Để biểu thức \(Q\) có giá trị âm thì \(\dfrac{3\sqrt{x}}{\sqrt{x}-2}< 0\)

\(\Rightarrow\sqrt{x}-2< 0\) (vì \(3\sqrt{x}>0\forall x>0;x\ne4\))

\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0\le x< 4\) 

Kết hợp với điều kiện xác định của \(x\), ta được: \(0< x< 4\)

\(\text{#}\mathit{Toru}\)

a: ĐKXĐ: x>0; x<>1

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

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

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

b: Q>2

=>Q-2>0

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

=>căn x-1>0

=>x>1

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

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

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

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

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

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

b) Q>2 <=> \(\dfrac{x}{\sqrt{x}-1}>2\Leftrightarrow x>2\sqrt{x}-2\)

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

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

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2\ge0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1\le0\\\sqrt{x}-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le1\end{matrix}\right.\)

KL:.....

a) \(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4,9\right)\)

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

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

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

b) \(\sqrt{x}=\sqrt{6+4\sqrt{2}}=\sqrt{\left(2+\sqrt{2}\right)^2}=2+\sqrt{2}\)

\(\Rightarrow Q=\dfrac{2+\sqrt{2}+1}{2+\sqrt{2}-3}=\dfrac{3+\sqrt{2}}{\sqrt{2}-1}=\dfrac{\left(3+\sqrt{2}\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)

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

c) \(Q=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)

Để \(Q\in Z\Rightarrow4⋮\sqrt{x}-3\Rightarrow\sqrt{x}-3\in\left\{1;2;4;-1;-2;-4\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{4;5;7;2;1\right\}\Rightarrow x\in\left\{16;25;49;4;1\right\}\)

a) Ta có: \(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

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

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

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

b) Thay \(x=6+4\sqrt{2}\) vào Q, ta được:

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

a) \(Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\) (ĐK: \(x\ne4,x\ge0\))

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

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

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

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

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

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

b) \(Q=\dfrac{6}{5}\) khi:

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

\(\Leftrightarrow15=12+6\sqrt{x}\)

\(\Leftrightarrow6\sqrt{x}=3\)

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

\(\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)