Câu hỏi
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 \(\begin{array}{l}
a)\,\,A = - \sqrt {3\left( {y - x} \right)} \\
b)\,\,\,B = 5\sqrt {3x} \\
c)\,\,C = -x\sqrt 3 \\
d)\,\,\,D = \sqrt {xy} \left( {10 + 3xy} \right)
\end{array}\)
Lời giải chi tiết:
\(a)\,\,\,A = \left( {x - y} \right)\sqrt {\frac{3}{{y - x}}} \)
Điều kiện: \(\frac{3}{{y - x}} > 0 \Leftrightarrow y - x > 0 \Leftrightarrow x < y \Rightarrow x - y < 0.\)
\( \Rightarrow A = \left( {x - y} \right)\sqrt {\frac{3}{{y - x}}} = - \sqrt {\frac{{3{{\left( {y - x} \right)}^2}}}{{y - x}}} = - \sqrt {3\left( {y - x} \right)} .\)
\(\begin{array}{l}b)\,\,B = 2\sqrt {3x} - \sqrt {48x} + \sqrt {108x} + \sqrt {3x} \,\,\,\,\left( {x \ge 0} \right)\\\,\,\,\,\,\,\,\,\,\,\,\, = 2\sqrt {3x} - \sqrt {{4^2}.3x} + \sqrt {{6^2}.3x} + \sqrt {3x} \\\,\,\,\,\,\,\,\,\,\,\,\, = 2\sqrt {3x} - 4\sqrt {3x} + 6\sqrt {3x} + \sqrt {3x} \\\,\,\,\,\,\,\,\,\,\,\,\, = 5\sqrt {3x} .\,\end{array}\)
\(\begin{array}{l}c)\,\,C = \frac{1}{{1 - 5x}}\sqrt {3{x^2}\left( {25{x^2} - 10x + 1} \right)} ,\,\,\,\,0 \le x \le \frac{1}{5}\\\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{1 - 5x}}\sqrt {3{x^2}{{\left( {5x - 1} \right)}^2}} = \frac{{\sqrt 3 \left| x \right|\left| {5x - 1} \right|}}{{1 - 5x}}\end{array}\)
Vì \(x \ge 0 \Rightarrow \left| x \right| = x;\,\,\,0 \le x \le \frac{1}{5} \Rightarrow 5x - 1 \le 0 \Rightarrow \left| {5x - 1} \right| = 1 - 5x.\)
\( \Rightarrow C = \frac{{\sqrt 3 x\left( {1 - 5x} \right)}}{{1 - 5x}} = \sqrt 3 x.\)
\(\begin{array}{l}d)\,\,D = 2\sqrt {25xy} + \sqrt {225{x^3}{y^3}} - 3y\sqrt {16{x^3}y} \,\,\,\,\left( {x \ge 0,\,\,y \ge 0} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\sqrt {{5^2}xy} + \sqrt {{{15}^2}{x^3}{y^3}} - 3y\sqrt {{4^2}{x^3}y} \\\,\,\,\,\,\,\,\,\,\,\,\,\, = 10\sqrt {xy} + 15\left| {xy} \right|\sqrt {xy} - 12\left| x \right|y\sqrt {xy} \\\,\,\,\,\,\,\,\,\,\,\,\,\, = 10\sqrt {xy} + 15xy\sqrt {xy} - 12xy\sqrt {xy} \,\,\,\,\left( {do\,\,\,x \ge 0,\,\,y \ge 0} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 10\sqrt {xy} + 3xy\sqrt {xy} \\\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {xy} \left( {10 + 3xy} \right).\end{array}\)
Câu hỏi trước
