Đưa thừa số vào trong dấu căn

\(\begin{array}{l}a)\,\,\frac{1}{{xy}}\sqrt {\frac{{{x^2}{y^2}}}{2}} & & & b)\,\,a\sqrt 2 & & & c)\,\, - \frac{a}{b}\sqrt {\frac{b}{a}} \,\,\,\left( {a > 0,\,\,\,b > 0} \right)\\d)\,\,a\sqrt {\frac{3}{a}} & & & e)\,\,\frac{1}{{2x - 1}}\sqrt {5\left( {1 - 4x + 4{x^2}} \right)} .\end{array}\)

Câu 130676: 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. \(\begin{array}{l}
a)\,\,\left\{ \begin{array}{l}
\frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy > 0\\
- \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy < 0
\end{array} \right.\\
b)\,\,\left\{ \begin{array}{l}
\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\
- \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0
\end{array} \right.\\
c)\,\,\sqrt {\frac{a}{b}} \\d)\,\, \sqrt{3} a\\
e)\,\,\left\{ \begin{array}{l}
\sqrt 5 \,\,\,khi\,\,\,x > \frac{1}{2}\\
- \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2}
\end{array} \right.
\end{array}\)

B. \(\begin{array}{l}
a)\,\,\left\{ \begin{array}{l}
\frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy > 0\\
- \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy < 0
\end{array} \right.\\
b)\,\,\left\{ \begin{array}{l}
\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\
- \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0
\end{array} \right.\\
c)\,\, - \sqrt {\frac{a}{b}} \\d)\,\, \sqrt{3a} \\
e)\,\, \sqrt{3} a\\e)\,\,\left\{ \begin{array}{l}\sqrt 5 \,\,\,khi\,\,\,x \ge \frac{1}{2}\\- \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2}\end{array} \right.\end{array}\)

C. \(\begin{array}{l}
a)\,\,\frac{{\sqrt 2 }}{2}\\
b)\,\,\left\{ \begin{array}{l}
\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\
- \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0
\end{array} \right.\\
c)\,\,\sqrt {\frac{a}{b}} \\d)\,\, \sqrt{3a} \\
e)\,\,\left\{ \begin{array}{l}
\sqrt 5 \,\,\,khi\,\,\,x \ge \frac{1}{2}\\
- \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2}
\end{array} \right.
\end{array}\)

D. \(\begin{array}{l}
a)\,\,\frac{{\sqrt 2 }}{2}\\
b)\,\,\left\{ \begin{array}{l}
\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\
- \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0
\end{array} \right.\\
c)\,\,\sqrt {\frac{a}{b}} \\d)\,\, \sqrt{3} a\\
e)\,\,\sqrt 5
\end{array}\)

Câu hỏi : 130676

Đáp án : B
(16) bình luận (0) lời giải
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: \(a\sqrt {\frac{3}{a}} = \sqrt {{a^2}.\frac{3}{a}} = \sqrt {3a} .\)

\(e)\,\,\frac{1}{{2x - 1}}\sqrt {5\left( {1 - 4x + 4{x^2}} \right)} = \left\{ \begin{array}{l}
\sqrt {\frac{1}{{{{\left( {2x - 1} \right)}^2}}}.5{{\left( {1 - 2x} \right)}^2}} \,\,\,khi\,\,\,2x - 1 > 0\\
- \sqrt {\frac{1}{{{{\left( {2x - 1} \right)}^2}}}.5{{\left( {1 - 2x} \right)}^2}} \,\,\,\,khi\,\,\,2x - 1 < 0
\end{array} \right. = \left\{ \begin{array}{l}
\sqrt 5 \,\,\,khi\,\,\,x > \frac{1}{2}\\
- \sqrt 5 \,\,\,\,khi\,\,\,x < \frac{1}{2}\,\,\,
\end{array} \right..\)

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