\(\hept{\begin{cases}\left(x+\sqrt{x^2+2021}\right)\left(y+\sqrt{y^2+2021}\right)\left(x-\sqrt{x^2+2021}\right)=\left(x-\sqrt{x^2+2021}\right)2021\\\left(x+\sqrt{x^2+2021}\right)\left(y+\sqrt{y^2+2021}\right)\left(y-\sqrt{y^2+2021}\right)=\left(y-\sqrt{y^2+2021}\right)2021\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}-2021\left(y+\sqrt{y^2+2021}\right)=\left(x-\sqrt{x^2+2021}\right)2021\\-2021\left(x+\sqrt{x^2+2021}\right)=\left(y-\sqrt{y^2+2021}\right)2021\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}y+\sqrt{y^2+2021}=\sqrt{x^2+2021}-x\\x+\sqrt{x^2+2021}=\sqrt{y^2+2021}-y\end{cases}}\)

\(\Rightarrow y+\sqrt{y^2+2021}+x+\sqrt{x^2+2021}=\sqrt{x^2+2021}-x+\sqrt{y^2+2021}-y\)

\(\Rightarrow x+y=0\)

a, Với a ; b > 0 

\(A=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)

b, Với x ; y > 0 ; \(x\ne y\)

\(B=\left(\frac{x\sqrt{x}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}+\sqrt{xy}\right):\left(x-y\right)-\frac{2\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

\(=\left(\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\sqrt{xy}\right):\left(x-y\right)-\frac{2\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

\(=\frac{2\sqrt{xy}}{x-y}-\frac{2\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{2\sqrt{xy}-2\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{x-y}=\frac{-2y}{x-y}\)

a, Với \(x\ge0;x\ne2\)

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

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

\(=\frac{3x+3\sqrt{x}-3-x+1-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{x-\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

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

b, \(\frac{\sqrt{x}-3}{\sqrt{x}-1}=\frac{\sqrt{x}-1-2}{\sqrt{x}-1}=1-\frac{2}{\sqrt{x}-1}\)

\(\Rightarrow\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

1, ĐK\(x\ge2\)

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

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

\(\Rightarrow\sqrt{x-2}-2+\sqrt{x-2}-3=5\)

\(\Rightarrow2\sqrt{x-2}=10\Rightarrow\sqrt{x-2}=5\Rightarrow x-2=25\Rightarrow x=27\)(TMĐK)

\(\hept{\begin{cases}x-my=0\\mx-y=m+1\end{cases}\hept{\begin{cases}x=my\left(1\right)\\mx-y=m+1\left(2\right)\end{cases}}}\)

thế (1) vào (2) ta được

\(m^2y-y=m+1\)

\(y\left(m^2-1\right)=m+1\)

\(y\left(m-1\right)\left(m+1\right)=m+1\)

nếu \(a\ne0\Rightarrow\left(m-1\right)\left(m+1\right)\ne0\Rightarrow m\ne\pm1\)

thì pt có nghiệm duy nhất là 

\(y=\frac{m+1}{\left(m-1\right)\left(m+1\right)}=\frac{1}{m-1}\)

thay \(y=\frac{1}{m-1}\)vào pt(1) ta đc

\(x-\frac{m}{m-1}=0\)

\(x=\frac{m}{m-1}\)

nếu \(a=0\Rightarrow\left(m-1\right)\left(m+1\right)=0\Rightarrow\orbr{\begin{cases}m=1\\m=-1\end{cases}}\)

nếu m=1

\(0a=m+1\Rightarrow0a=2\)(vô lí) vậy hpt vô nghiệm với m=1

nếu m=-1

\(0a=m+1\Rightarrow0a=-1+1=0\)(luôn đúng) vậy hpt vô số nghiệm với m=-1

ta có :

\(\widehat{OAB}+\widehat{O'AC}=90^o\Rightarrow\hept{\begin{cases}AC=2AO\cos\widehat{OAC}\\AB=2AO'\cos\widehat{O'AB}=2AO'\sin\widehat{OAC}\end{cases}}\)

ta có : \(S_{ABC}=\frac{1}{2}AB.AC=2OA.O'A.\sin\widehat{OAC}.cos\widehat{OAC}\le OA.O'A\left(\sin^2\widehat{OAC}+cos^2\widehat{OAC}\right)=OA.OA'\)

dấu bằng xảy ra khi \(\sin\widehat{OAC}=cos\widehat{OAC}\Rightarrow\widehat{OAC}=45^o\)

từ đó ta xác định được vị trí của B và C

điều kiện xác định :

\(\hept{\begin{cases}x\left(x-1\right)\ge0\\x\left(x+2\right)\ge0\\x^2\ge0\end{cases}}\) với \(x\left(x-1\right)\ge0\Leftrightarrow\orbr{\begin{cases}x\ge1\\x\le0\end{cases}}\)

với \(x\left(x+2\right)\ge0\Leftrightarrow\orbr{\begin{cases}x\ge0\\x\le-2\end{cases}}\) còn \(x^2\ge0\) luôn đúng

Kết hợp điều kiện ta có :

\(\orbr{\begin{cases}x\le-2\\x\ge1\end{cases}}\) hoặc \(x=0\)