Tìm miền xác định phải không 

a) 

\(1-\sqrt{2x-x^2}\) 

a xác định \(\Leftrightarrow2x-x^2\ge0\) 

\(0\le x\le2\) 

b) 

\(\sqrt{-4x^2+4x-1}\) 

b xác định 

\(\Leftrightarrow-4x^2+4x-1\ge0\) 

\(-\left(4x^2-4x+1\right)\ge0\) 

\(4x^2-4x+1\le0\) 

\(\left(2x-1\right)^2\le0\) 

2x - 1 = 0 

x = 1/2 

c) 

\(\frac{x}{\sqrt{5x^2-3}}\) 

c xác định 

\(\Leftrightarrow5x^2-3>0\) 

\(5x^2>3\) 

\(x^2>\frac{3}{5}\) 

\(\orbr{\begin{cases}x< -\frac{\sqrt{15}}{5}\\x>\frac{\sqrt{15}}{5}\end{cases}}\) 

d) 

d xác định 

\(\Leftrightarrow\sqrt{x-\sqrt{2x-1}}>0\) 

\(x-\sqrt{2x-1}>0\) 

\(x>\sqrt{2x-1}\) 

\(\hept{\begin{cases}2x-1\ge0\\x^2>2x-1\end{cases}}\) 

\(\hept{\begin{cases}x\ge\frac{1}{2}\\x^2-2x+1>0\end{cases}}\) 

\(\hept{\begin{cases}x\ge\frac{1}{2}\\\left(x-1\right)^2>0\end{cases}}\) 

\(\hept{\begin{cases}x\ge\frac{1}{2}\\x-1\ne0\end{cases}}\) 

\(\hept{\begin{cases}x\ge\frac{1}{2}\\x\ne1\end{cases}}\) 

e) 

e xác định 

\(\Leftrightarrow\frac{-2x^2}{3x+2}\ge0\) 

\(3x+2< 0\) ( vì \(-2x^2\le0\forall x\) ) 

\(x< -\frac{2}{3}\) 

f) 

f xác định 

\(\Leftrightarrow x^2+x-2>0\) 

\(\orbr{\begin{cases}x< -2\\x>1\end{cases}}\)

= 0 bạn nhé

\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)

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

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

\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)

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

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

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

Vì \(9>5\)\(\Rightarrow\sqrt{9}>\sqrt{5}\)\(\Rightarrow3>\sqrt{5}\)\(\Rightarrow3-\sqrt{5}>0\)

mà \(3+\sqrt{5}>0\)

\(\Rightarrow\left(3-\sqrt{5}\right).\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right).\sqrt{3-\sqrt{5}}\)

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

\(=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)

\(=\sqrt{\left(9-5\right)\left(3-\sqrt{5}\right)}+\sqrt{\left(9-5\right).\left(3+\sqrt{5}\right)}\)

\(=\sqrt{4.\left(3-\sqrt{5}\right)}+\sqrt{4.\left(3+\sqrt{5}\right)}\)

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

\(\sqrt{n^2+n^2\left(n+1\right)^2+\left(n+1\right)^2}\)

\(=\sqrt{n^2+\left(n^2+n\right)^2+\left(n^2+2n+1\right)}\)

\(=\sqrt{2\left(n^2+n\right)+\left(n^2+n\right)^2+1}\)

\(=\sqrt{\left(n^2+n+1\right)^2}=\left|n^2+n+1\right|=n^2+n+1\)

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