Toán Lớp 9: tìm đkxđ của biểu thức
1) 1/ √x+2
2) √3/ -2 + x
3) √x+1 / x
4) 2/ √1 – x
5) 1/ √x – 1/ √x – 2
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Toán Lớp 9: tìm đkxđ của biểu thức
1) 1/ √x+2
2) √3/ -2 + x
3) √x+1 / x
4) 2/ √1 – x
5) 1/ √x – 1/ √x – 2
Comments ( 2 )
1)\dfrac{1}{{\sqrt {x + 2} }}\\
Dkxd:x + 2 > 0\\
\Leftrightarrow x > – 2\\
Vậy\,x > – 2\\
2)\sqrt {\dfrac{3}{{ – 2 + x}}} \\
Dkxd:\dfrac{3}{{ – 2 + x}} \ge 0\\
\Leftrightarrow – 2 + x > 0\\
\Leftrightarrow x > 2\\
Vậy\,x > 2\\
3)\sqrt {\dfrac{{x + 1}}{x}} \\
Dkxd:\dfrac{{x + 1}}{x} \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 \ge 0\\
x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 \le 0\\
x < 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x > 0\\
x \le – 1
\end{array} \right.\\
Vậy\,x \le – 1\,hoặc\,x > 0\\
4)\dfrac{2}{{\sqrt {1 – x} }}\\
Dkxd:1 – x > 0\\
\Leftrightarrow x < 1\\
Vậy\,x < 1\\
5)\dfrac{1}{{\sqrt x }} – \dfrac{1}{{\sqrt x – 2}}\\
Dkxd:\left\{ \begin{array}{l}
x > 0\\
\sqrt x \# 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x\# 4
\end{array} \right.\\
Vậy\,x > 0;x\# 4
\end{array}$