Với \(x\ge0;x\ne9\)

\(B=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)

\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}=\frac{3x+9\sqrt{x}}{x-9}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)

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

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

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

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

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

\(7+\sqrt{2x-3}=1+\sqrt{2x-3}\)(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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\(\sqrt{225}\) = 15

HT

a, \(\sqrt{x^2-6x+9}=4\Leftrightarrow\sqrt{\left(x-3\right)^2}=4\Leftrightarrow\left|x-3\right|=4\)

TH1 : \(x-3=4\Leftrightarrow x=7\)

TH2 : \(x-3=-4\Leftrightarrow x=-1\)

b, \(\sqrt{x^2+2x+1}=3\Leftrightarrow\sqrt{\left(x+1\right)^2}=3\Leftrightarrow\left|x+1\right|=3\)

TH1 : \(x+1=3\Leftrightarrow x=2\)

TH2 : \(x+1=-3\Leftrightarrow x=-4\)

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

TH1 : \(2x-1=5\Leftrightarrow x=3\)

TH2 : \(2x-1=-5\Leftrightarrow x=-2\)