a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)

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

Thay x = 4 => \(\sqrt{x}=2\) vào B ta được : 

\(B=\frac{2+5}{2-3}=-7\)

b, Ta có : Với \(x\ge0;x\ne9\)

\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)

\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\)

Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)

K=\(\frac{\sqrt{x}+1}{\sqrt{x}+3}+\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{2x-10}{x+2\sqrt{x}-3}ĐK:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)-2x+10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

=\(\frac{x-1-2x+3\sqrt{x}-2\sqrt{x}-1-6+10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

Để K>0 thì :\(\frac{1}{\sqrt{x}-1}>0\Leftrightarrow\sqrt{x}-1>0\Leftrightarrow x>1\)

Với x>1 thoả mãn yêu cầu.

\(A=\frac{2a-3\sqrt{a}-2}{\sqrt{a}-2}\\ =\frac{2a-4\sqrt{a}+\sqrt{a}-2}{\sqrt{a}-2}\\ =\frac{\left(2\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\\ =2\sqrt{a}+1\)

\(M=a-\sqrt{a-2016}\\ =a-2016-2.\frac{1}{2}.\sqrt{a-2016}+\frac{1}{4}+2015,75\)

\(=\left(\sqrt{a-2016}-\frac{1}{2}\right)^2+2015,75\)

\(M=a-\sqrt{a-2016}\\ =a-\sqrt{a-2016}-2016+2016\\ =a-2016-\sqrt{a-2016}+2016\)

Bài 1 : 

a )\(A=\frac{3-\sqrt{3}}{\sqrt{3}-1}+\frac{\sqrt{35}-\sqrt{15}}{\sqrt{5}}-\sqrt{28}\)

\(A=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\frac{\sqrt{5}\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{5}}-\sqrt{28}\)

\(A=\sqrt{3}+\sqrt{7}-\sqrt{3}-\sqrt{28}\)

\(A=\sqrt{7}-\sqrt{28}\)

\(A=\sqrt{7}-2\sqrt{7}=-\sqrt{7}\)

Vậy \(A=-\sqrt{7}\)

b)\(B=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\left(a,b>0;a\ne b\right)\)

\(B=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\)

\(B=\left(\sqrt{a}+\sqrt{b}\right).\frac{a-b}{\sqrt{a}+\sqrt{b}}\)

\(B=a-b\)

Vậy \(B=a-b\left(a,b>0;a\ne b\right)\)

_Minh ngụy_

Bài 2 :

a )\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\left(x>0\right)\)

\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\frac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\frac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

Vậy \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)

b) \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)

Ta có : \(B>0\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}>0\)

Vì : \(\sqrt{x}\ge0\forall x\Rightarrow\)để \(B>O\)cần \(\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)( thỏa mãn \(x>0\))

Vậy \(x>1\)thì \(B>0\)

_Minh ngụy_

Bài 3: 

a) \(K=\frac{5+\sqrt{5}}{\sqrt{5}+1}-\sqrt{6-2\sqrt{5}}=\frac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=\sqrt{5}-\left|\sqrt{5}-1\right|=1\)

KL: K=1

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

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

\(=\frac{-4\sqrt{x}}{\left(1-x\right)}\cdot\left(1-x\right)x\sqrt{x}=-4x^2\)

b) Thay P = -4x^2 và K= 1 vào biểu thức P + 6K =2x , được: 

\(-4x^2+6=2x\Leftrightarrow2x^2+x-6=0\Leftrightarrow\left(2x-3\right)\left(x+2\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\left(n\right)\\x=-2\left(n\right)\end{cases}}\)

KL:.......