\(A=\sqrt{\sqrt{3}-1}\left(\sqrt{6}+\sqrt{2}\right)\)

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

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

\(A=\sqrt{3-1}.\sqrt{\sqrt{3}+1}.\sqrt{2}\)

\(A=\sqrt{2}.\sqrt{2}.\sqrt{\sqrt{3}+1}=2.\sqrt{\sqrt{3}+1}\)

Vậy \(A=2\sqrt{\sqrt{3}+1}\).

Với x = 9 tmdk thay vào A ta được : \(A=\frac{\sqrt{9}+3}{9-4}=\frac{3+3}{5}=\frac{6}{5}\)

\(B=\frac{-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{1}{\sqrt{x}+2}\)

\(P=\frac{A}{B}=\frac{\frac{\sqrt{x}+3}{x-4}}{\frac{1}{\sqrt{x}+2}}=\frac{\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+2}{1}=\frac{\sqrt{x}+3}{\sqrt{x}-2}\)

Xét hiệu P - 1 ta có : \(P-1=\frac{\sqrt{x}+3}{\sqrt{x}-2}-1=\frac{\sqrt{x}+3-\sqrt{x}+2}{\sqrt{x}-2}=\frac{5}{\sqrt{x}-2}>0\forall x>4\)

=> P > 1

a, Thay x = 9 vào A ta được : \(A=\frac{3+3}{9-4}=\frac{6}{5}\)

b, Với \(x\ge0;x\ne4\)

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

c, với x > 4 Ta có : \(P=\frac{A}{B}\Rightarrow\frac{\sqrt{x}+3}{x-4}:\frac{1}{\sqrt{x}+2}=\frac{\sqrt{x}+3}{\sqrt{x}-2}\)

Ta có : \(1=\frac{\sqrt{x}-2}{\sqrt{x}-2}\)mà \(\sqrt{x}+3>\sqrt{x}-2\)

Vậy P > 1 

S = \(\frac{1}{B}+A\)=>  \(\frac{1}{\frac{\sqrt{x}}{\sqrt{x}-3}}+\frac{x+7}{\sqrt{x}}=\frac{\sqrt{x}-3+x+7}{\sqrt{x}}=\frac{x+\sqrt{x}+4}{\sqrt{x}}\)

\(=\sqrt{x}+1+\frac{4}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{4}{\sqrt{x}}}+1=5\)

Dấu ''='' xảy ra khi x = 4 

b, \(\sqrt{\left(2-\sqrt{3}\right)^2}+\frac{2}{\sqrt{3}+1}-6\sqrt{\frac{16}{3}}\)

\(=2-\sqrt{3}+\frac{2\left(\sqrt{3}-1\right)}{2}-\frac{6.4}{\sqrt{3}}\)

\(=2-\sqrt{3}+\sqrt{3}-1-\frac{24\sqrt{3}}{3}=1-8\sqrt{3}\)

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

\(4+\sqrt{2x-3-6\sqrt{2x-3}+9}=\sqrt{2x-3+2\sqrt{2x-3}+1}\)

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

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

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

\(7+\sqrt{2x-3}=1+\sqrt{2x-3}\)(vô lý)

pt vô nghiệm

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lại nhầm nữa sr

bạn sủa dòng 3 thành

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

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

\(TH1:x\le6\)

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

\(7-\sqrt{2x-3}=-\sqrt{2x-3}-1\)

\(7=-1\)vô nghiệm
\(TH2:x>6\)

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

\(\sqrt{2x-3}+1=\sqrt{2x-3}+1\)pt vô số nghiệm

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