Tính:a) \(\dfrac{x}{y}\sqrt {\dfrac{{{x^2}}}{{{y^4}}}} {\mkern 1mu} {\mkern 1mu} \left( {x > 0,y \ne 0} \right)\)b)
Tính:
a) \(\dfrac{x}{y}\sqrt {\dfrac{{{x^2}}}{{{y^4}}}} {\mkern 1mu} {\mkern 1mu} \left( {x > 0,y \ne 0} \right)\)
b) \(5xy\sqrt {\dfrac{{25{x^2}}}{{{y^6}}}} {\mkern 1mu} {\mkern 1mu} \left( {x < 0,y > 0} \right)\)
c) \(a{b^2}\sqrt {\dfrac{3}{{{a^2}{b^4}}}} {\mkern 1mu} {\mkern 1mu} \left( {a < 0} \right)\)
d) \(\sqrt {\dfrac{{9 + 12a + 4{a^2}}}{{{b^2}}}} {\mkern 1mu} {\mkern 1mu} \left( {a \ge - \dfrac{3}{2},{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} b < 0} \right)\)
Áp dụng phép khai phương một thương và phép khai phương một tích.
a) \(\dfrac{x}{y} \cdot \sqrt {\dfrac{{{x^2}}}{{{y^4}}}} {\rm{\;}}\,\,(x > 0,y \ne 0)\)
\( = \dfrac{x}{y} \cdot \dfrac{{\sqrt {{x^2}} }}{{\sqrt {{y^4}} }} = \dfrac{x}{y} \cdot \dfrac{{\left| x \right|}}{{{y^2}}} = \dfrac{x}{y} \cdot \dfrac{x}{{{y^2}}} = \dfrac{{{x^2}}}{{{y^3}}}{\rm{\;}}\,\,({\rm{do\;}}x > 0)\)
b) \(5xy\sqrt {\dfrac{{25{x^2}}}{{{y^6}}}} {\mkern 1mu} {\mkern 1mu} \left( {x < 0,y > 0} \right)\)
\( = 5xy \cdot \dfrac{{\sqrt {25{x^2}} }}{{\sqrt {{y^6}} }} = 5xy \cdot \dfrac{{\sqrt {{{(5x)}^2}} }}{{\sqrt {{{\left( {{y^3}} \right)}^2}} }}\)
\(\; = 5xy \cdot \dfrac{{\left| {5x} \right|}}{{\left| {{y^3}} \right|}} = 5xy \cdot \dfrac{{ - 5x}}{{{y^3}}} = \dfrac{{ - 25{x^2}}}{{{y^2}}}{\rm{\;}}\,\,({\rm{do\;}}x > 0;y < 0)\)
c) \(a{b^2}\sqrt {\dfrac{3}{{{a^2}{b^4}}}} {\mkern 1mu} {\mkern 1mu} \left( {a < 0} \right)\)
\( = a{b^2} \cdot \dfrac{{\sqrt 3 }}{{\sqrt {{a^2} \cdot {b^4}} }} = a{b^2} \cdot \dfrac{{\sqrt 3 }}{{\left| {a{b^2}} \right|}} = a{b^2} \cdot \dfrac{{\sqrt 3 }}{{ - a{b^2}}} = - \sqrt 3 {\rm{\;}}\,\,\left( {a < 0 \Rightarrow \left| {a{b^2}} \right| = - a{b^2}} \right)\)
d) \(\sqrt {\dfrac{{9 + 12a + 4{a^2}}}{{{b^2}}}} {\mkern 1mu} {\mkern 1mu} \left( {a \ge - \dfrac{3}{2},{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} b < 0} \right)\)
\(\; = \sqrt {\dfrac{{{{(3 + 2a)}^2}}}{{{b^2}}}} = \dfrac{{\sqrt {{{(3 + 2a)}^2}} }}{{\sqrt {{b^2}} }} = \dfrac{{\left| {3 + 2a} \right|}}{{\left| b \right|}} = \dfrac{{3 + 2a}}{{ - b}}\) \(\left( {{\rm{do\;}}a \ge - \dfrac{3}{2} \Rightarrow 2a + 3 \ge 0;b < 0} \right)\)
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