\(A=\dfrac{4}{2-x}+\dfrac{100}{x}+2021=36\left(2-x\right)+\dfrac{4}{2-x}+36x+\dfrac{100}{x}+1949\)

\(0< x< 2\Rightarrow\left\{{}\begin{matrix}x>0\\x< 2\Rightarrow-x>-2\Leftrightarrow2-x>0\end{matrix}\right.\)

\(\Rightarrow A\ge2\sqrt{36\left(2-x\right).\dfrac{4}{\left(2-x\right)}}+2\sqrt{36x.\dfrac{100}{x}}+1985=2\sqrt{4.36}+2\sqrt{36.100}+1949=2093\Rightarrow A_{min}=2093\Leftrightarrow\left\{{}\begin{matrix}36\left(2-x\right)=\dfrac{4}{2-x}\\36x=\dfrac{100}{x}\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{5}{3}\left(tm\right)\)

a) \(ĐK:x\ge0,x\ne1\)

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

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

b) \(P=\dfrac{2\sqrt{x}}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\)

Kết hợp với đk:

\(\Rightarrow0\le x< 1\)

\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\\ A\in Z\Rightarrow1+\dfrac{2}{\sqrt{x}-1}\in Z\Rightarrow\dfrac{2}{\sqrt{x}-1}\in Z\\ \Leftrightarrow\left(\sqrt{x}-1\right)\inƯ\left(2\right)\\ \Leftrightarrow\left(\sqrt{x}-1\right)\in\left\{2;1;-1;-2\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{3;2;0;-1\right\}\\ \Leftrightarrow x\in\left\{9;4;0\right\}\)

Vậy \(x\in\left\{9;4;0\right\}\)

\(a.P=\dfrac{x\sqrt{x}-47}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}-\dfrac{4\sqrt{x}+12}{\sqrt{x}+1}+\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=\dfrac{x\sqrt{x}-47-4\left(x-9\right)+\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\dfrac{x\sqrt{x}-3x+3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(x+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\dfrac{x+3}{\sqrt{x}+1}\left(x\ne9;x\ge0\right)\)

\(b.P=\dfrac{x+3}{\sqrt{x}+1}=\dfrac{x-1+4}{\sqrt{x}+1}=\sqrt{x}-1+\dfrac{4}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2\)

Áp dụng BĐT Cauchy cho các số dương , ta có :

\(\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}\ge2\sqrt{\left(\sqrt{x}+1\right).\dfrac{4}{\sqrt{x}+1}}=2\sqrt{4}=4\)

\(\Leftrightarrow\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2\ge4-2=2\)

\(\Rightarrow P_{Min}=2."="\Leftrightarrow x=1\left(TM\right)\)

\(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\x\ne9\\\end{matrix}\right.\\ P=\dfrac{x\sqrt{x}+3\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\\ P=\dfrac{\sqrt{x}\left(x+3\right)-3\left(x+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\\ P=\dfrac{x+3}{\sqrt{x}+1}\)

b) \(P=\dfrac{x+3}{\sqrt{x}+1}=\dfrac{2\left(\sqrt{x}+1\right)+\left(x-2\sqrt{x}+1\right)}{\sqrt{x}+1}\\ P=2+\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\\ =>P\ge2\\ Dấubằngxảyrakhix=1\)

\(A=\dfrac{x-4+16}{\sqrt{x}+2}=\sqrt{x}+2+\dfrac{16}{\sqrt{x}+2}-4\)

\(\Leftrightarrow A\ge2\cdot\sqrt{16}-4=2\cdot4-4=4\)

Dấu '=' xảy ra khi \(\sqrt{x}+2=4\)

hay x=4

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)

\(M=A\cdot B=\dfrac{x}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)

=>\(M=\dfrac{x}{\sqrt{x}+2}\)

=>\(M=\dfrac{x-4+4}{\sqrt{x}+2}=\sqrt{x}-2+\dfrac{4}{\sqrt{x}+2}\)

=>\(M=\sqrt{x}+2+\dfrac{4}{\sqrt{x}+2}-4\)

=>\(M>=2\cdot\sqrt{\left(\sqrt{x}+2\right)\cdot\dfrac{4}{\sqrt{x}+2}}-4=0\)

Dấu '=' xảy ra khi \(\sqrt{x}+2=\sqrt{4}=2\)

=>\(\sqrt{x}=0\)

=>x=0(nhận)