Tháng Sáu 5, 2023

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

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..\)