a: \(2+\frac{x-4}{\sqrt{x}+2}=2+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=2+\sqrt{x}-2=\sqrt{x}\)
\(\frac{2}{\sqrt{x}+3}-\frac{1}{\sqrt{x}+1}+\frac{4}{x+4\sqrt{x}+3}\)
\(=\frac{2}{\sqrt{x}+3}-\frac{1}{\sqrt{x}+1}+\frac{4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2\left(\sqrt{x}+1\right)-\sqrt{x}-3+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\frac{2\sqrt{x}+2-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\frac{1}{\sqrt{x}+1}\)
ĐKXĐ: x>=0
Ta có: \(B=\left(2+\frac{x-4}{\sqrt{x}+2}\right):\left(\frac{2}{\sqrt{x}+3}-\frac{1}{\sqrt{x}+1}+\frac{4}{x+4\sqrt{x}+3}\right)\)
\(=\sqrt{x}:\frac{1}{\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}+1\right)\)
b: Ta có: \(\left(2+B+2+3\right)\left(2^2+B^2+2^2+3^2\right)^2=2023\)
=>\(\left(B+7\right)\left(B^2+17\right)^2=2023\)
=>\(\left(B+7\right)\left(B^4+34B^2+289\right)=2023\)
=>\(B^5+34B^3+289B+7B^4+238B^2+2023=2023\)
=>\(B^5+7B^4+34B^3+238B^2+289B=0\)
=>\(B\left(B^4+7B^3+34B^2+238B-289\right)=0\)
=>B=0
=>\(\sqrt{x}\left(\sqrt{x}+1\right)=0\)
=>\(\sqrt{x}=0\)
=>x=0
