a: Khi x=16 thì \(A=\dfrac{6}{16-3\cdot4}=\dfrac{6}{4}=\dfrac{3}{2}\)

b: P=A:B

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

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

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

c: \(P-1=\dfrac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}}=\dfrac{3}{\sqrt{x}}>0\)

=>P>1

\(M=\left(\dfrac{3}{\sqrt{x}+3}+\dfrac{x+9}{x-9}\right):\left(\dfrac{2\sqrt{x}-5}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)

\(=\dfrac{3\sqrt{x}-9+x+9}{x-9}:\dfrac{2\sqrt{x}-5-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

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

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

Lời giải:
ĐKXĐ: $x\geq -2$

PT $\Leftrightarrow 2\sqrt{x+2}+3\sqrt{4}.\sqrt{x+2}-\sqrt{9}.\sqrt{x+2}=10$

$\Leftrightarrow 2\sqrt{x+2}+6\sqrt{x+2}-3\sqrt{x+2}=10$

$\Leftrightarrow 5\sqrt{x+2}=10$

$\Leftrightarrow \sqrt{x+2}=2$

$\Leftrightarrow x+2=4$

$\Leftrightarrow x=2$ (tm)

1: =>x^2-x=3-x

=>x^2=3

=>x=căn 3 hoặc x=-căn 3

2: =>x^2-4x+3=x^2-4x+4 và x>=2

=>3=4(vô lý)

3: =>2|x-1|=6

=>|x-1|=3

=>x-1=3 hoặc x-1=-3

=>x=-2 hoặc x=4

4: =>|2x-3|=|x-2|

=>2x-3=x-2 hoặc 2x-3=-x+2

=>x=1 hoặc x=5/3

5: =>\(\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)=0\)

=>x+2=0

=>x=-2

\(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+2}-\dfrac{3\sqrt{x}}{x+\sqrt{x}-2}\left(x\ge0\right)\\ A=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\left(\sqrt{x}-1\right)-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\\ A=\dfrac{x+2\sqrt{x}+\sqrt{x}-1-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\\ A=\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

Lần sau ghi đề rõ ra nha bạn

a) \(3x-2\sqrt{x-1}=4\) (ĐK: x ≥ 1)

\(\Rightarrow3x-2\sqrt{x-1}-4=0\)

\(\Rightarrow3x-6-2\sqrt{x-1}+2=0\)

\(\Rightarrow3\left(x-2\right)-2\left(\sqrt{x-1}-1\right)=0\)

\(\Rightarrow3\left(x-2\right)-2.\dfrac{x-2}{\sqrt{x-1}+1}=0\)

\(\Rightarrow\left(x-2\right)\left[3-\dfrac{2}{\sqrt{x-1}+1}\right]=0\)

*TH1: x = 2 (t/m)

*TH2: \(3-\dfrac{2}{\sqrt{x-1}+1}=0\)

\(\Rightarrow3=\dfrac{2}{\sqrt{x-1}+1}\)

\(\Rightarrow3\sqrt{x-1}+3=2\)

\(\Rightarrow3\sqrt{x-1}=-1\) (vô lí)

Vậy S = {2}

b) \(\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\) (ĐK: \(-\dfrac{1}{4}\le x\le3\) )

\(\Rightarrow\sqrt{4x+1}-3-\sqrt{x+2}+2-\sqrt{3-x}+1=0\)

\(\Rightarrow\dfrac{4x-8}{\sqrt{4x+1}+3}-\dfrac{x-2}{\sqrt{x+2}+2}+\dfrac{x-2}{\sqrt{3-x}+1}=0\)

\(\Rightarrow\left(x-2\right)\left(\dfrac{4}{\sqrt{4x+1}+3}-\dfrac{1}{\sqrt{x+2}+2}+\dfrac{1}{\sqrt{3-x}+1}\right)=0\)

=> x = 2

\(a,3x-2\sqrt{x-1}=4\left(x\ge1\right)\\ \Leftrightarrow-2\sqrt{x-1}=4-3x\\ \Leftrightarrow4\left(x-1\right)=16-24x+9x^2\\ \Leftrightarrow9x^2-28x+20=0\\ \Leftrightarrow\left(x-2\right)\left(9x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=\dfrac{10}{9}\left(tm\right)\end{matrix}\right.\)

\(b,\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\left(-\dfrac{1}{4}\le x\le3\right)\\ \Leftrightarrow4x+1+x+2-2\sqrt{\left(4x+1\right)\left(x+2\right)}=3-x\\ \Leftrightarrow-2\sqrt{\left(4x+1\right)\left(x+2\right)}=2-6x\\ \Leftrightarrow\sqrt{4x^2+9x+2}=3x-1\\ \Leftrightarrow4x^2+9x+2=9x^2-6x+1\\ \Leftrightarrow5x^2-15x-1=0\\ \Leftrightarrow\Delta=225+20=245\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15-\sqrt{245}}{10}=\dfrac{15-7\sqrt{5}}{10}\left(ktm\right)\\x=\dfrac{15+\sqrt{245}}{10}=\dfrac{15+7\sqrt{5}}{10}\left(tm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{15+7\sqrt{5}}{10}\)

\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}}{x-9}+\dfrac{3x+3}{x-9}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}+1}\)

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

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

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

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

Q=√x−2+√x+2=2√x