by Tính:
\(\begin{array}{l}
a)\,\,\frac{x}{y}\sqrt {\frac{{{x^2}}}{{{y^4}}}} \,\,\left( {x > 0,y \ne 0} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)\,\,5xy\sqrt {\frac{{25{x^2}}}{{{y^6}}}} \,\,\left( {x < 0,y > 0} \right)\\
c)\,\,a{b^2}\sqrt {\frac{3}{{{a^2}{b^4}}}} \,\,\left( {a < 0} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,d)\,\,\sqrt {\frac{{9 + 12a + 4{a^2}}}{{{b^2}}}} \,\,\left( {a \ge – \frac{3}{2},\,\,\,b < 0} \right)
\end{array}\)
A \(\begin{array}{l}
a)\,\,\frac{{{x^2}}}{{{y^3}}} & & & b)\,\, \frac{{25{x^2}}}{y}\\
c)\,\, – \sqrt 3 & & d)\,\,\frac{{3 + 2a}}{{ – b}}
\end{array}\)
B \(\begin{array}{l}
a)\,\,\frac{{{x^2}}}{{{y^3}}} & & & b)\,\, – \frac{{25{x^2}}}{y}\\
c)\,\, – \sqrt 3 & & d)\,\,\frac{{3 + 2a}}{{ – b}}
\end{array}\)
C \(\begin{array}{l}
a)\,\,\frac{{{x^2}}}{{{y^3}}} & & & b)\,\, – \frac{{25{x^2}}}{y}\\
c)\,\, \sqrt 3 & & d)\,\,\frac{{3 + 2a}}{{ – b}}
\end{array}\)
D \(\begin{array}{l}
a)\,\,\frac{{{x^2}}}{{{y^3}}} & & & b)\,\, – \frac{{25{x^2}}}{y}\\
c)\,\, -\sqrt 3 & & d)\,\,\frac{{3 + 2a}}{{ b}}
\end{array}\)
Hướng dẫn Chọn đáp án là: B
Lời giải chi tiết:
\(\begin{array}{l}a)\,\,\frac{x}{y}.\sqrt {\frac{{{x^2}}}{{{y^4}}}} \,\,\,\left( {x > 0,\,\,y \ne 0} \right) = \frac{x}{y}.\frac{{\sqrt {{x^2}} }}{{\sqrt {{y^4}} }} = \frac{x}{y}.\frac{{\left| x \right|}}{{{y^2}}} = \frac{x}{y}.\frac{x}{{{y^2}}} = \frac{{{x^2}}}{{{y^3}}}\,\,\,\left( {do\,\,\,x > 0} \right)\\b)\,\,5xy.\sqrt {\frac{{25{x^2}}}{{{y^6}}}} \,\,\left( {x < 0,\,\,y > 0} \right) = 5xy.\frac{{\sqrt {25{x^2}} }}{{\sqrt {{y^6}} }} = 5xy.\frac{{\sqrt {{{\left( {5x} \right)}^2}} }}{{\sqrt {{{\left( {{y^3}} \right)}^2}} }}\\ = 5xy.\frac{{\left| {5x} \right|}}{{\left| {{y^3}} \right|}} = 5xy.\frac{{ – 5x}}{{{y^3}}} = \frac{{ – 25{x^2}}}{y}\,\,\,\,\,\left( {do\,\,x > 0;\,\,y < 0} \right).\\c)\,\,a{b^2}.\sqrt {\frac{3}{{{a^2}{b^4}}}} = a{b^2}.\frac{{\sqrt 3 }}{{\sqrt {{a^2}.{b^4}} }} = a{b^2}.\frac{{\sqrt 3 }}{{\left| {a{b^2}} \right|}} = a{b^2}.\frac{{\sqrt 3 }}{{ – a{b^2}}} = – \sqrt 3 \,\,\,\left( {a < 0 \Rightarrow \left| {a{b^2}} \right| = – a{b^2}} \right).\\d)\,\,\sqrt {\frac{{9 + 12a + 4{a^2}}}{{{b^2}}}} \,\,\,\,\left( {a \ge – \frac{3}{2};\,\,b < 0} \right)\\ = \sqrt {\frac{{{{\left( {3 + 2a} \right)}^2}}}{{{b^2}}}} = \frac{{\sqrt {{{\left( {3 + 2a} \right)}^2}} }}{{\sqrt {{b^2}} }} = \frac{{\left| {3 + 2a} \right|}}{{\left| b \right|}} = \frac{{3 + 2a}}{{ – b}}\\\left( {do\,\,\,a \ge – \frac{3}{2} \Rightarrow 2a + 3 \ge 0;\,\,b < 0} \right).\end{array}\)
