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Giải câu h,k,l, m, n dùm
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RuslanHeatt
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Đáp án:
$\begin{array}{l}
h)\sqrt {{x^2} – 4x} + 4\\
Dkxd:{x^2} – 4x \ge 0\\
\Rightarrow x\left( {x – 4} \right) \ge 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 0\\
x – 4 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 0\\
x – 4 \le 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x \ge 4\\
x \le 0
\end{array} \right.\\
Vay\,x \ge 4\,hoac\,x \le 0\\
k)\sqrt {{x^2} – 4x + 4} – 1\\
Dkxd:{x^2} – 4x + 4 \ge 0\\
\Rightarrow {\left( {x – 2} \right)^2} \ge 0\dforall x\\
l)\sqrt { – {x^2} + 6x + 14} \ge 0\\
Dkxd: – {x^2} + 6x + 14 \ge 0\\
\Rightarrow {x^2} – 6x – 14 \le 0\\
\Rightarrow {x^2} – 6x + 9 \le 23\\
\Rightarrow {\left( {x – 3} \right)^2} \le 23\\
\Rightarrow – \sqrt {23} + 3 \le x \le \sqrt {23} + 3\\
m)Dkxd:\left\{ \begin{array}{l}
x – 1 \ge 0\\
x – x \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ge 1\\
0 \ne 0
\end{array} \right.\\
\Rightarrow x \in \emptyset \\
n)Dkxd:\left\{ \begin{array}{l}
2x – 1 + \dfrac{1}{{\sqrt {x – 3} }} \ge 0\\
x – 3 > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2x + \dfrac{1}{{\sqrt {x – 3} }} \ge 1\\
x > 3
\end{array} \right.\\
Khi:x > 3\\
\Rightarrow 2x > 6\\
\Rightarrow 2x + \dfrac{1}{{\sqrt {x – 3} }} > 6 > 1\left( {tm} \right)\\
Vay\,x > 3
\end{array}$
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