a) \(\sqrt{x-3}\) xác định
\(\Leftrightarrow x-3\ge0\)
\(\Leftrightarrow x\ge3\)
Vậy..
b) \(\sqrt{3-2x}\) xác định
\(\Leftrightarrow3-2x\ge0\)
\(\Leftrightarrow x\le-\dfrac{3}{2}\)
Vậy..
c) \(\sqrt{4x^2-1}\) xác định
\(\Leftrightarrow4x^2-1\ge0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-1\ge0\\2x+1\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ge\dfrac{-1}{2}\end{matrix}\right.\)\(\Rightarrow x\ge\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}2x-1\le0\\2x+1\le0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x\le\dfrac{-1}{2}\end{matrix}\right.\) \(\Rightarrow x\le\dfrac{-1}{2}\)
Vậy ...
d) \(\sqrt{3x^2+2}\) xác định
\(\Leftrightarrow3x^2+2\ge0\)
mà \(3x^2\ge0\)
\(\Rightarrow3x^2+2>0\)
Vậy...
e) \(\sqrt{2x^2+4x+5}\) xác định
\(\Leftrightarrow2x^2+4x+5\ge0\)
mà \(2x^2+4x\ge0\)
\(2x\left(x+2\right)\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}2x\ge0\\x+2\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-2\end{matrix}\right.\)\(\Rightarrow x\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}2x\le0\\x+2\le0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\le0\\x\le-2\end{matrix}\right.\)\(\Rightarrow x\le-2\)
\(\Rightarrow2x^2+4x+5>0\)
Vậy...
( Câu này không chắc lắm nha )
Bài 2: Tách sẵn ra cho bạn luôn nhé, không thì bạn nhấn máy tính ra cũng được :v
a) \(-\dfrac{7}{9}\sqrt{\left(-27\right)^2+6\sqrt{1}}\)
\(=-\dfrac{7}{9}\sqrt{\left(-3\right)^2.\left(-9\right)^2+6}\)
\(=\dfrac{-7}{9}\sqrt{735}\)
\(=\dfrac{-7}{9}\sqrt{49.15}\)
\(=\dfrac{-49\sqrt{15}}{9}\)
b) \(\sqrt{49}\sqrt{12^2}+\sqrt{256}:\sqrt{8^2}\)
\(=84+2=86\)
c)\(\sqrt{\left(\sqrt{3-1}\right)^2-\sqrt{\left(\sqrt{3+1}\right)^2}}\)
\(=\sqrt{2-2}\)
= 0
