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

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

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

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

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

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

Bài 1:

\(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{5+2\sqrt{5}+1}\)

\(=\left|4-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=4-\sqrt{5}+\sqrt{5}+1=5\)

Bài 2:

a: ĐKXĐ: x>=3

\(\sqrt{x-3}=6\)

=>x-3=36

=>x=36+3=39(nhận)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(x-3\right)^2}=12\)

=>\(\left|x-3\right|=12\)

=>\(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)

Bài 3:

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

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

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

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

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

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

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

c: \(A=\sqrt{3x-1}+3\cdot\sqrt{12x-4}-\sqrt{6^2\left(3x-1\right)}+\sqrt{5}\)

\(=\sqrt{3x-1}+6\sqrt{3x-1}-6\sqrt{3x-1}+\sqrt{5}\)

\(=\sqrt{3x-1}+\sqrt{5}\)

d: \(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)

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

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

\(=\dfrac{2\left(a-2\right)}{a+2}\)

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

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

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

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

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

a:

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

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

b: Khi x=7-4căn 3 thì 

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

c: A=3

=>căn x-2=1

=>x=9(loại)

\(a,A=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\left(dkxd:x\ne4,x\ge0,x\ne9\right)\)

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

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

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

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

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

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

\(b,x=7-4\sqrt{3}\Rightarrow A=\dfrac{3}{\sqrt{7-4\sqrt{3}}-2}=\dfrac{3}{\sqrt{\left(\sqrt{3}-2\right)^2}-2}=\dfrac{3}{\left|\sqrt{3}-2\right|-2}=\dfrac{3}{-\sqrt{3}+2-2}=\dfrac{\sqrt{3^2}}{-\sqrt{3}}=-\sqrt{3}\)

\(c,A=3\Rightarrow\dfrac{3}{\sqrt{x}-2}=3\\ \Rightarrow\dfrac{3-3\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=0\\ \Rightarrow3-3\sqrt{x}+6=0\\ \Rightarrow-3\sqrt{x}=-9\\ \Rightarrow\sqrt{x}=3\\ \Rightarrow x=9\left(ktm\right)\)

Vậy không có giá trị x thỏa mãn đề bài.

\(1,ĐKx\ge5\)

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

\(\Rightarrow\sqrt{x-5}\left(\sqrt{x+5}+2\right)-3\left(\sqrt{x+5}+2\right)=0\)

\(\Rightarrow\left(\sqrt{x+5}+2\right)\left(\sqrt{x-5}-3\right)=0\)

\(\left[{}\begin{matrix}\sqrt{x+5}=-2loại\\\sqrt{x-5}=3\end{matrix}\right.\)\(\Rightarrow x-5=9\Rightarrow x=14\)(TMĐK)

2a,ĐK \(x\ge0;x\ne9\)

,\(B=\dfrac{7\left(3-\sqrt{x}\right)-12}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}\)

\(M=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

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

`a)M=(sqrtx/(sqrtx-3)+sqrtx/(sqrtx+3))*(x-9)/sqrt{9x}(x>0,x ne 9)`

`M=((sqrtx(sqrtx+3)+sqrtx(sqrtx-3))/(x-9))*(x-9)/(3sqrtx)`

`M=((x+3sqrtx+x-3sqrtx)/(x-9))*(x-9)/(3sqrtx)`

`M=(2x)/(3sqrtx)=(2sqrtx)/3`

`b)M=6`

`=>2sqrtx=18`

`=>sqrtx=9=>x=81(tmđk)`

a) \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+3}\right).\dfrac{x-9}{\sqrt{9x}}\left(x>0,x\ne9\right)\)

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

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

b) \(M=6\Rightarrow\dfrac{2\sqrt{x}}{3}=6\Rightarrow2\sqrt{x}=18\Rightarrow\sqrt{x}=9\Rightarrow x=81\)

a) Ta có: \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+3}\right)\cdot\dfrac{x-9}{\sqrt{9x}}\)

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

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

b) Để M=6 thì \(\dfrac{2}{3}\sqrt{x}=6\)

\(\Leftrightarrow\sqrt{x}=9\)

hay x=81(thỏa ĐK)

a) ĐK: x ≥ 0; x ≠ 9; x≠4

P= \(\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{3}{x-5\sqrt{x}+6}\right):\left(\dfrac{x+2}{\sqrt{x}-3}-\dfrac{x^2-\sqrt{x}-6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\right)\)

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

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

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

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

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

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

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

b)  P ≤ -2

⇒ \(\dfrac{x^2-3x+2}{x-4}\) ≤ -2

⇔ \(\dfrac{x^2-3x+2}{x-4}\) + 2 ≤ 0

⇔ \(\dfrac{x^2-3x+2+2\left(x-4\right)}{x-4}\) ≤ 0

⇔ \(\dfrac{x^2-3x+2+2x-8}{x-4}\) ≤ 0

⇔\(\dfrac{x^2-x-6}{x-4}\) ≤ 0

⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-x-6\ge0\\x-4< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-x-6\le0\\x-4>0\end{matrix}\right.\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x\le2\\3\le x< 4\end{matrix}\right.\)

Vậy.......

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:

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