Giải thích các bước giải:
Bài 1:
Các biểu thức đã cho có nghĩa khi và chỉ khi:
\(\begin{array}{l}
1,\\
- 3x + 4 \ge 0 \Leftrightarrow 3x - 4 \le 0 \Leftrightarrow 3x \le 4 \Leftrightarrow x \le \dfrac{4}{3}\\
2,\\
- 5x - 25 \ge 0 \Leftrightarrow 5x + 25 \le 0 \Leftrightarrow 5x \le - 25 \Leftrightarrow x \le - 5\\
3,\\
\dfrac{{x + 3}}{7} \ge 0 \Leftrightarrow x + 3 \ge 0 \Leftrightarrow x \ge - 3\\
4,\\
\left\{ \begin{array}{l}
\dfrac{{ - 5}}{{4x - 22}} \ge 0\\
4x - 22 \ne 0
\end{array} \right. \Leftrightarrow 4x - 22 < 0 \Leftrightarrow 4x < 22 \Leftrightarrow x < \dfrac{{11}}{2}\\
5,\\
2{x^2} + 3 \ge 0,\,\,\,\forall x
\end{array}\)
Bài 2:
Ta có:
\(\begin{array}{l}
a,\\
\sqrt {6,{8^2} - 3,{2^2}} = \sqrt {\left( {6,8 - 3,2} \right)\left( {6,8 + 3,2} \right)} = \sqrt {3,6.10} = \sqrt {36} = 6\\
b,\\
\sqrt {21,{8^2} - 18,{2^2}} = \sqrt {\left( {21,8 - 18,2} \right)\left( {21,8 + 18,2} \right)} = \sqrt {3,6.40} = \sqrt {36.4} = \sqrt {{6^2}{{.2}^2}} = 6.2 = 12\\
c,\\
\sqrt {\dfrac{9}{{169}}} = \sqrt {\dfrac{{{3^2}}}{{{{13}^2}}}} = \dfrac{3}{{13}}\\
d,\\
\sqrt {\dfrac{{25}}{{144}}} = \sqrt {\dfrac{{{5^2}}}{{{{12}^2}}}} = \sqrt {{{\left( {\dfrac{5}{{12}}} \right)}^2}} = \dfrac{5}{{12}}\\
e,\\
\sqrt {1\dfrac{9}{{16}}} = \sqrt {\dfrac{{16 + 9}}{{16}}} = \sqrt {\dfrac{{25}}{{16}}} = \sqrt {{{\left( {\dfrac{5}{4}} \right)}^2}} = \dfrac{5}{4}
\end{array}\)
