\(A=\sqrt{\dfrac{18-3\sqrt{3}}{11}}-\sqrt{2+\sqrt{3}}\)

\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{3}+1}{\sqrt{2}}\)

\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{6}+\sqrt{2}}{2}\)

\(=\dfrac{2\sqrt{11\left(18-3\sqrt{3}\right)}-11\sqrt{6}-11\sqrt{2}}{22}\)

b: \(=\dfrac{x\sqrt{x}-2x+28-x+16-x-9\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\sqrt{x}-4x-9\sqrt{x}+36}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}=\dfrac{x-9}{\sqrt{x}+1}\)

a: \(A=\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\dfrac{4+\sqrt{3}}{5-2\sqrt{3}}}\)

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

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

\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}\)

b: \(B=\dfrac{x\sqrt{x}-2x+28}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}-\dfrac{x-16}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}-\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\sqrt{x}-2x+28-x+16-x-9\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\sqrt{x}-4\sqrt{x}-9\sqrt{x}+36}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}=\dfrac{x-9}{\sqrt{x}+1}\)

Đặt \(\sqrt{x-1}=a\), khi đó ta có:

\(P=\left(\dfrac{\sqrt{x-1}}{3+\sqrt{x-1}}+\dfrac{x+8}{10-x}\right):\left(\dfrac{3\sqrt{x-1}+1}{x-3\sqrt{x-1}-1}-\dfrac{1}{\sqrt{x-1}}\right)\)

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

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

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

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

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

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

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

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

Ta lại có: \(x=\sqrt[4]{\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}}-\sqrt[4]{\dfrac{3-2\sqrt{2}}{3+2\sqrt{2}}}\)

\(=\sqrt[4]{\dfrac{\left(\sqrt{2}+1\right)^2}{\left(\sqrt{2}-1\right)^2}}-\sqrt[4]{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}+1\right)^2}}\)

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

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

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

\(=2\).

Suy ra: \(P=\dfrac{-3\sqrt{2-1}}{2\left(\sqrt{2-1}+2\right)}=\dfrac{-3}{2.3}=-\dfrac{1}{2}\).

b, ĐKXĐ: \(x\ge\frac{5}{2}\)

\(pt\Leftrightarrow\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}+1\right)^2}=4\)

\(\Leftrightarrow\sqrt{2x-5}=3\)

\(\Leftrightarrow x=7\left(tm\right)\)

a, ĐKXĐ: \(x\ge5\)

\(pt\Leftrightarrow\sqrt{x-5+4\sqrt{x-5}+4}+\sqrt{x-5+8\sqrt{x-5}+16}=0\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-5}+2\right)^2}+\sqrt{\left(\sqrt{x-5}+4\right)^2}=0\)

\(\Leftrightarrow2\sqrt{x-5}+6=0\)

\(\Leftrightarrow\sqrt{x-5}=-3\)

Phương trình vô nghiệm

c: \(=\sqrt{\dfrac{4}{16-6\sqrt{7}}}+\sqrt{7}\)

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

\(=3+2\sqrt{7}\)

d: \(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{x-4}\)

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

ta có : \(A=\dfrac{x+y-2\sqrt{xy}}{\sqrt{x}-\sqrt{y}}-\dfrac{x+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)

\(=\sqrt{x}-\sqrt{y}-\sqrt{x}=-\sqrt{y}\)

ta có \(B=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\dfrac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)

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

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

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

\(\Rightarrow\sqrt{x}+3\)

\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)

\(\Rightarrow\sqrt{y}-1\)

\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)

\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)

\(\Rightarrow\sqrt{xy}\)

\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)

\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)

\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)

\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)

\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)

\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)

\(f,\sqrt{8-2\sqrt{15}}+\sqrt{5}+\sqrt{3}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}+\sqrt{5}+\sqrt{3}\)

\(\Leftrightarrow\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}\)

\(\Rightarrow2\sqrt{5}-\sqrt{3}+\sqrt{3}\)

\(g,\sqrt{9-2\sqrt{4}}-\sqrt{9+2\sqrt{14}}\)

\(\Leftrightarrow\sqrt{9-2\times2}-\sqrt{\left(\sqrt{2}+\sqrt{7}\right)^2}\)

\(\Leftrightarrow\sqrt{9-4}-\left(\sqrt{2}+\sqrt{7}\right)\)

\(\Rightarrow\sqrt{5}-\sqrt{2}-\sqrt{7}\)

1)

ĐK: \(x\geq 5\)

PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)

\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)

\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)

\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)

2)

ĐK: \(x\geq -1\)

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

\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)

\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)

\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)

\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)

Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$

\(\Rightarrow x=3\) (thỏa mãn)

Vậy .............

3)

ĐK: \(x^2-6x+7\geq 0\)

Đặt \(\sqrt{x^2-6x+7}=a(a\geq 0)\) \(\Rightarrow x^2-6x=a^2-7\)

PT trở thành: \(a^2-7+a=5\Leftrightarrow a^2+a-12=0\)

\(\Leftrightarrow (a-3)(a+4)=0\Rightarrow a=3\) (do \(a\geq 0)\)

\(\Rightarrow \sqrt{x^2-6x+7}=3\)

\(\Rightarrow x^2-6x+7=9\)

\(\Leftrightarrow x^2-6x-2=0\) \(\Rightarrow x=3\pm \sqrt{11}\) (đều thỏa mãn)