by Đưa thừa số vào trong dấu căn
\(\begin{array}{l}a)\,\,\frac{1}{{xy}}\sqrt {\frac{{{x^2}{y^2}}}{2}} & & & b)\,\,a\sqrt 2 & & & c)\,\, – \frac{a}{b}\sqrt {\frac{b}{a}} \,\,\,\left( {a > 0,\,\,\,b > 0} \right)\\d)\,\,a\sqrt {\frac{3}{a}} & & & e)\,\,\frac{1}{{2x – 1}}\sqrt {5\left( {1 – 4x + 4{x^2}} \right)} .\end{array}\)
A \(\begin{array}{l}
a)\,\,\left\{ \begin{array}{l}
\frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy > 0\\
– \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy < 0
\end{array} \right.\\
b)\,\,\left\{ \begin{array}{l}
\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\
– \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0
\end{array} \right.\\
c)\,\,\sqrt {\frac{a}{b}} \\
d)\,\,\left\{ \begin{array}{l}
\sqrt 5 \,\,\,khi\,\,\,x > \frac{1}{2}\\
– \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2}
\end{array} \right.
\end{array}\)
B \(\begin{array}{l}
a)\,\,\left\{ \begin{array}{l}
\frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy > 0\\
– \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy < 0
\end{array} \right.\\
b)\,\,\left\{ \begin{array}{l}
\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\
– \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0
\end{array} \right.\\
c)\,\, – \sqrt {\frac{a}{b}} \\
d)\,\,\left\{ \begin{array}{l}
\sqrt 5 \,\,\,khi\,\,\,x \ge \frac{1}{2}\\
– \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2}
\end{array} \right.
\end{array}\)
C \(\begin{array}{l}
a)\,\,\frac{{\sqrt 2 }}{2}\\
b)\,\,\left\{ \begin{array}{l}
\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\
– \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0
\end{array} \right.\\
c)\,\,\sqrt {\frac{a}{b}} \\
d)\,\,\left\{ \begin{array}{l}
\sqrt 5 \,\,\,khi\,\,\,x \ge \frac{1}{2}\\
– \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2}
\end{array} \right.
\end{array}\)
D \(\begin{array}{l}
a)\,\,\frac{{\sqrt 2 }}{2}\\
b)\,\,\left\{ \begin{array}{l}
\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\
– \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0
\end{array} \right.\\
c)\,\,\sqrt {\frac{a}{b}} \\
d)\,\,\sqrt 5
\end{array}\)
Hướng dẫn Chọn đáp án là: B
Lời giải chi tiết:
\(\begin{array}{l}a)\,\,\frac{1}{{xy}}\sqrt {\frac{{{x^2}{y^2}}}{2}} = \left\{ \begin{array}{l}\frac{{\sqrt 2 }}{2}.\sqrt {\frac{{{x^2}{y^2}}}{{{x^2}{y^2}}}} \,\,\,khi\,\,\,xy > 0\\ – \frac{{\sqrt 2 }}{2}.\sqrt {\frac{{{x^2}{y^2}}}{{{x^2}{y^2}}}} \,\,\,khi\,\,\,xy < 0\end{array} \right. = \left\{ \begin{array}{l}\frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy > 0\\ – \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy < 0\end{array} \right..\\b)\,\,\,a\sqrt 2 = \left\{ \begin{array}{l}\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\ – \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0\end{array} \right..\\c)\,\, – \frac{a}{b}\sqrt {\frac{b}{a}} \,\,\,\,\left( {a > 0,\,\,b > 0} \right)\end{array}\)
Ta có: \(a > 0;\,\,b > 0 \Rightarrow \frac{a}{b} > 0\)
\( \Rightarrow – \frac{a}{b}\sqrt {\frac{b}{a}} = – \sqrt {{{\left( {\frac{a}{b}} \right)}^2}\frac{b}{a}} = – \sqrt {\frac{a}{b}} .\)
\(d)\,\,a\sqrt {\frac{3}{a}} \)
Điều kiện: \(a > 0.\)
Ta có: \(a\sqrt {\frac{3}{a}} = \sqrt {{a^2}.\frac{3}{a}} = \sqrt {3a} .\)
\(e)\,\,\frac{1}{{2x – 1}}\sqrt {5\left( {1 – 4x + 4{x^2}} \right)} = \left\{ \begin{array}{l}
\sqrt {\frac{1}{{{{\left( {2x – 1} \right)}^2}}}.5{{\left( {1 – 2x} \right)}^2}} \,\,\,khi\,\,\,2x – 1 > 0\\
– \sqrt {\frac{1}{{{{\left( {2x – 1} \right)}^2}}}.5{{\left( {1 – 2x} \right)}^2}} \,\,\,\,khi\,\,\,2x – 1 < 0
\end{array} \right. = \left\{ \begin{array}{l}
\sqrt 5 \,\,\,khi\,\,\,x > \frac{1}{2}\\
– \sqrt 5 \,\,\,\,khi\,\,\,x < \frac{1}{2}\,\,\,
\end{array} \right..\)
