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

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

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

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

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

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

b: Để P=20 thì \(\sqrt{x}+2=60x-80\sqrt{x}\)

\(\Leftrightarrow60x-81\sqrt{x}-2=0\)

Đặt \(\sqrt{x}=a\)

Pt sẽ là \(60a^2-81a-2=0\)

\(\text{Δ}=\left(-81\right)^2-4\cdot60\cdot\left(-2\right)=7041>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}a_1=\dfrac{81-\sqrt{7041}}{120}\left(loại\right)\\a_2=\dfrac{81+\sqrt{7041}}{120}\left(nhận\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\left(\dfrac{81+\sqrt{7041}}{120}\right)^2\)

a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

\(A=\dfrac{x^2-x^2-2x+2x-4}{\left(x-2\right)\left(x+2\right)}\left(x\ne\pm2\right)=\dfrac{-4}{x^2-4}\\ B=\dfrac{\left|x-1\right|+x^2+4x+4}{\left(x-2\right)\left(x+2\right)}\left(x\ne\pm2\right)\)

Với \(x>1;x\ne2\Leftrightarrow B=\dfrac{x^2+5x+3}{\left(x-2\right)\left(x+2\right)}\)

Với \(x< 1;x\ne-2\Leftrightarrow B=\dfrac{x^2+3x+5}{\left(x-2\right)\left(x+2\right)}\)

\(\dfrac{x+4}{x^2-4}-\dfrac{2}{x^2+2x}\)

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

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

đk : x >= 0 ; x khác 4 

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

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

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

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

ĐKXĐ: x khác 4; x ≥ 0

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

a) ĐKXĐ: 

\(\left\{{}\begin{matrix}x^2-9\ne0\\x+3\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne-3\end{matrix}\right.\Leftrightarrow x\ne\pm3\) 

b) \(A=\dfrac{x+15}{x^2-9}-\dfrac{2}{x+3}\)

\(A=\dfrac{x+15}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{x+15-2x+6}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{21-x}{\left(x+3\right)\left(x-3\right)}\)

c) Thay x = - 1 vào A ta có: 

\(A=\dfrac{21-\left(-1\right)}{\left(-1+3\right)\left(-1-3\right)}=\dfrac{21+1}{2\cdot-4}=\dfrac{22}{-8}=-\dfrac{11}{4}\)