Biến đổi đơn giản biểu thức chứa căn bậc hai
Câu hỏi 1 :
Đưa thừa số ra ngoài dấu căn:\(a)\,\,\,\sqrt {180{x^2}} \\ b)\,\sqrt {3{x^2} - 6xy + 3{y^2}} \)
A \(\begin{array}{l}
a)\,\,6x\sqrt 5 \\
b)\,\,\,\left\{ \begin{array}{l}
\left( {x - y} \right)\sqrt 3 \,\,\,\,khi\,\,\,\,x \ge y\\
\left( {y - x} \right)\sqrt x \,\,\,\,khi\,\,\,x < y
\end{array} \right..
\end{array}\)
B \(\begin{array}{l}
a)\,\,\, - 6x\sqrt 5 \\
b)\,\,\left( {x - y} \right)\sqrt 3
\end{array}\)
C \(\begin{array}{l}
a)\,\,\left\{ \begin{array}{l}
6x\sqrt 5 \,\,\,khi\,\,\,x \ge 0\\
- 6x\sqrt 5 \,\,\,\,khi\,\,\,x < 0
\end{array} \right.\\
b)\,\,\,\left\{ \begin{array}{l}
\left( {x - y} \right)\sqrt 3 \,\,\,khi\,\,\,x \ge 0\\
\left( {y - x} \right)\sqrt 3 \,\,\,khi\,\,\,x < y
\end{array} \right..
\end{array}\)
D \(\begin{array}{l}
a)\,\, - 6x\sqrt 5 \\
b)\,\, - \left( {x - y} \right)\sqrt 3
\end{array}\)
Câu hỏi 2 :
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 3 :
Rút gọn:\(\begin{array}{l}a)\,\,\,A = \left( {x - y} \right)\sqrt {\frac{3}{{y - x}}} \\b)\,\,B = 2\sqrt {3x} - \sqrt {48x} + \sqrt {108x} + \sqrt {3x} \,\,\,\,\,\left( {x \ge 0} \right)\\c)\,\,\,C = \frac{1}{{1 - 5x}}\sqrt {3{x^2}\left( {25{x^2} - 10x + 1} \right)} ,\,\,\,0 \le x \le \frac{1}{5}\\d)\,\,D = 2\sqrt {25xy} + \sqrt {225{x^3}{y^3}} - 3y\sqrt {16{x^3}y} \,\,\,\,\left( {x \ge 0,\,\,y \ge 0} \right).\end{array}\)
A \(\begin{array}{l}
a)\,\,A = - \sqrt {3\left( {y - x} \right)} \\
b)\,\,\,B = 5\sqrt {3x} \\
c)\,\,C = x\sqrt 3 \\
d)\,\,\,D = \sqrt {xy} \left( {10 + 3xy} \right)
\end{array}\)
B \(\begin{array}{l}
a)\,\,A = \sqrt {3\left( {y - x} \right)} \\
b)\,\,\,B = 12\sqrt {3x} \\
c)\,\,C = x\sqrt 3 \\
d)\,\,\,D = \sqrt {xy} \left( {10 + 3xy} \right)
\end{array}\)
C \(\begin{array}{l}
a)\,\,A = - \sqrt {3\left( {y - x} \right)} \\
b)\,\,\,B = 12\sqrt {3x} \\
c)\,\,C = x\sqrt 3 \\
d)\,\,\,D = - \sqrt {xy} \left( {10 + 3xy} \right)
\end{array}\)
D \(\begin{array}{l}
a)\,\,A = - \sqrt {3\left( {y - x} \right)} \\
b)\,\,\,B = 5\sqrt {3x} \\
c)\,\,C = -x\sqrt 3 \\
d)\,\,\,D = \sqrt {xy} \left( {10 + 3xy} \right)
\end{array}\)
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