\(A=\frac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\frac{2\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{\sqrt{x}+3}\left(Đk:x\ge0;x\ne1\right)\)

\(=\frac{x\sqrt{x}+26\sqrt{x}-19}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x\sqrt{x}+16\sqrt{x}-x-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x\left(\sqrt{x}-1\right)+16\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x+16}{\sqrt{x}+3}\)

Ta có:\(\frac{x+16}{\sqrt{x}+3}=\frac{x-9+25}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+25}{\sqrt{x}+3}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\)

Vì \(x>0\Rightarrow\sqrt{x}+3>0\)

Áp dụng BĐT cô-si cho hai số dương  \(\sqrt{x+3}\)và\(\frac{25}{\sqrt{x}+3}\)ta có:

\(\sqrt{x}+3+\frac{25}{\sqrt{x}+3}\ge2\sqrt{\left(\sqrt{x}+3\right).\frac{25}{\sqrt{x}+3}}\)

\(\Rightarrow A\ge4\)

\(\Rightarrow MinA=4\Leftrightarrow\sqrt{x}+3=\frac{25}{\sqrt{x}+3}\Leftrightarrow\left(\sqrt{x}+3\right)^2=25\Leftrightarrow x=4\left(TMĐK\right)\)

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

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

\(y=\sqrt{\left(\sqrt{6}-1\right)^2}=\sqrt{6}-1\)

\(\Rightarrow x-y=1\Rightarrow P=1\)

\(B=x-2020-\sqrt{x-2020}+\dfrac{1}{4}+\dfrac{8079}{4}\)

\(B=\left(\sqrt{x-2020}-\dfrac{1}{2}\right)^2+\dfrac{8079}{4}\ge\dfrac{8079}{4}\)

\(B_{min}=\dfrac{8079}{4}\) khi \(x=\dfrac{8081}{4}\)

\(A=\sqrt{x-2\sqrt{x-1}}\)\(+5\sqrt{x+3-4\sqrt{x-1}}\)\(+8\sqrt{x+8-6\sqrt{x-1}}\)

\(=\sqrt{x-1-2\sqrt{x-1}+1}\)\(+5\sqrt{x-1-4\sqrt{x-1}+4}\)\(+8\sqrt{x-1-6\sqrt{x-1}+9}\)

\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}\)\(+5\sqrt{\left(\sqrt{x-1}-2\right)^2}\)\(+8\sqrt{\left(\sqrt{x-1}-3\right)^2}\)

\(=\sqrt{x-1}-1+5\sqrt{x-1}-10+8\sqrt{x-1}-24\)

\(=16\sqrt{x-1}-35\)

\(A_{min}=-35\Leftrightarrow16\sqrt{x-1}=0\Rightarrow x=1\)

a: B(căn x+3)=10 căn x

=>x+16-10 căn x=0

=>(căn x-2)(căn x-8)=0

=>x=4 hoặc x=64

b: \(B=\dfrac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\)

=>\(B=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6>=2\cdot\sqrt{25}-6=2\cdot5-6=4\)

Dấu = xảy ra khi (căn x+3)^2=25

=>căn x+3=5

=>căn x=2

=>x=4

a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}-2-\sqrt{x}+3}\)

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

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