Question

Biểu thức sau đây xác định với giá trị nào của x?

a) \(\sqrt{\left(x-1\right)\left(x-3\right)}\)

b)\(\sqrt{x^2-4}\)

c)\(\sqrt{\frac{x-2}{x+3}}\)

d) \(\sqrt{\frac{2+x}{5-x}}\)

Answer 1

a)\(\sqrt{\left(x-1\right)\left(x-3\right)}\ge0\)

\(\Rightarrow\left(x-1\right)\left(x-3\right)\ge0\)

\(\Rightarrow1\le x\le3\)

b)\(\sqrt{x^2-4}\)

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

\(\Rightarrow\left(x-2\right)\left(x+2\right)\ge0\)

\(\Rightarrow-2\le x\le2\)

c)\(\sqrt{\frac{x-2}{x+3}}=\frac{\sqrt{x-2}}{\sqrt{x+3}}\)

\(\Rightarrow\sqrt{x-2}\ge0\)

\(\Rightarrow x\ge2\)

\(\Rightarrow\sqrt{x+3}>0\)

\(\Rightarrow x+3>0\Leftrightarrow x>-3\)

\(\Rightarrow x\in\left(-\infty;-3\right)\)U[\(2;\infty\))

d)\(\sqrt{\frac{2+x}{5-x}}=\frac{\sqrt{2+x}}{\sqrt{5-x}}\)

\(\Rightarrow\sqrt{2+x}\ge0\)

\(\Rightarrow2+x\ge0\)

\(\Rightarrow x\ge-2\)

\(\Rightarrow\sqrt{5-x}>0\)

\(\Rightarrow5-x>0\Leftrightarrow x>5\)

\(\Rightarrow x\in\)[-2;5)

 

 

 

 

Answer 2

a) ĐKXĐ : \(\left(x-1\right)\left(x-3\right)\ge0\Leftrightarrow\begin{cases}x-1\ge0\\x-3\ge0\end{cases}\)hoặc \(\begin{cases}x-1\le0\\x-3\le0\end{cases}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x\ge3\\x\le1\end{array}\right.\)

b) \(x^2-4\ge0\Leftrightarrow x^2\ge4\Leftrightarrow\left|x\right|\ge2\Leftrightarrow\left[\begin{array}{nghiempt}x\ge2\\x\le-2\end{array}\right.\)

c) \(\frac{x-2}{x+3}\ge0\Leftrightarrow\begin{cases}x-2\ge0\\x+3>0\end{cases}\) hoặc \(\begin{cases}x-2\le0\\x+3< 0\end{cases}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x\ge2\\x< -3\end{array}\right.\)

d) \(\frac{2+x}{5-x}\ge0\) \(\Leftrightarrow\begin{cases}2+x\ge0\\5-x>0\end{cases}\) hoặc \(\begin{cases}2+x\le0\\5-x< 0\end{cases}\)

\(\Leftrightarrow-2\le x< 5\)