\(\sqrt{\dfrac{a^3}{a}}\)   với a <0 ta đc kết quả là

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27 tháng 10 2020

\(A=\frac{x}{y}.\frac{x}{y^2}=\frac{x^2}{y^3}\left(\text{vì }x>0;y< 0\text{ nên: }\frac{x}{y^2}>0\right)\)

27 tháng 10 2020

\(A=\frac{x}{y}\cdot\sqrt{\frac{x^2}{y^4}}=\frac{x}{y}\cdot\frac{\sqrt{x^2}}{\sqrt{y^4}}=\frac{x}{y}\cdot\frac{\left|x\right|}{\left|y^2\right|}=\frac{x}{y}\cdot\frac{x}{y^2}=\frac{x^2}{y^3}\)( x > 0 ; y < 0 )

27 tháng 10 2020

\(\frac{\sqrt{a^3}}{\sqrt{a}}\left(a>0\right)\)

\(=\sqrt{\frac{a^3}{a}}\)

\(=\sqrt{a^2}\)

\(=a\) (vì a>0)

27 tháng 10 2020

\(\frac{\sqrt{a^3}}{\sqrt{a}}=\frac{\sqrt{a^2\cdot a}}{\sqrt{a}}=\frac{\left|a\right|\sqrt{a}}{\sqrt{a}}=\left|a\right|=a\)( vì a > 0 )

a,\(ab^2\sqrt{\dfrac{3}{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{ab^2}=\sqrt{3}\)

b,\(\sqrt{\dfrac{27\left(a-3\right)^2}{48}}=\dfrac{3\sqrt{3}\left(a-3\right)}{4\sqrt{3}}=\dfrac{3}{4}\left(a-3\right)\)

c,\(\sqrt{\dfrac{9+12a+4a^2}{b^2}}=\dfrac{\sqrt{\left(3+2a\right)^2}}{\sqrt{b^2}}=\dfrac{3+2a}{b}\)

d, \(\left(a-b\right).\sqrt{\dfrac{ab}{\left(a-b\right)^2}}=\left(a-b\right).\dfrac{\sqrt{ab}}{\sqrt{\left(a-b\right)^2}}=\left(a-b\right).\dfrac{\sqrt{ab}}{\left(a-b\right)}=\sqrt{ab}\)

24 tháng 4 2017

a. \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{2.3}{3^2}}=\dfrac{1}{3}.\sqrt{6}\)

b. \(\sqrt{\dfrac{x^2}{5}}=\sqrt{\dfrac{5x^2}{5^2}}=\dfrac{x}{5}.\sqrt{5}\) (vì x \(\ge\) 0)

c. \(\sqrt{\dfrac{3}{x}}=\sqrt{\dfrac{3.x}{x^2}}=\dfrac{1}{x}.\sqrt{3x}\) (vì x > 0)

d. \(\sqrt{x^2-\dfrac{x^2}{7}}=\sqrt{\dfrac{6x^2}{7}}=\sqrt{\dfrac{6x^2.7}{7.7}}=\sqrt{\dfrac{42.x^2}{7^2}}=-\dfrac{x}{7}.\sqrt{42}\) (vì x < 0)

a: \(=2ab\cdot\dfrac{-15}{b^2a}=\dfrac{-30}{b}\)

b: \(=\dfrac{2}{3}\cdot\left(1-a\right)=\dfrac{2}{3}-\dfrac{2}{3}a\)

c: \(=\dfrac{\left|3a-1\right|}{\left|b\right|}=\dfrac{3a-1}{b}\)

d: \(=\left(a-2\right)\cdot\dfrac{a}{-\left(a-2\right)}=-a\)

27 tháng 10 2020

\(A=\frac{y}{x}\cdot\sqrt{\frac{x^2}{y^4}}=\frac{y}{x}\cdot\frac{\sqrt{x^2}}{\sqrt{y^4}}=\frac{y}{x}\cdot\frac{\left|x\right|}{\left|y^2\right|}=\frac{y}{x}\cdot\frac{x}{y^2}=\frac{1}{y}\)( x > 0 ; y > 0 )

9 tháng 7 2018

a) \(\dfrac{\sqrt{63y^3}}{\sqrt{7y}}=\sqrt{\dfrac{63y^3}{7y}}=\sqrt{9y^2}=\left|3y\right|=3y\) (vì y > 0)

b) \(\dfrac{\sqrt{68a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{68a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{17}{32a^2}}\)

30 tháng 5 2017

a. \(\sqrt{\dfrac{63y^3}{7y}}\)=\(\sqrt{9y^2}\)=3y

b.\(\sqrt{\dfrac{48x^3}{3x^5}}\)=\(\sqrt{16\cdot\dfrac{1}{X^2}}\)= \(\sqrt{16}\cdot\sqrt{\dfrac{1}{X^2}}\)=\(4\cdot\dfrac{1}{X}=\dfrac{4}{X}\)

c.\(\sqrt{\dfrac{45mn^2}{20m}}=\sqrt{\dfrac{9n^2}{4}}=\dfrac{\sqrt{9n^2}}{\sqrt{4}}=\dfrac{3n}{2}\)

d. \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{2\sqrt{2}a}\)

19 tháng 9 2017

a) \(\dfrac{\sqrt{63y^3}}{\sqrt{7y}}=\sqrt{\dfrac{63y^3}{7y}}=\sqrt{9y^2}=3y\)

b) \(\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\dfrac{48x^3}{3x^5}}=\sqrt{\dfrac{16}{x^2}}=\dfrac{4}{x}\)

c) \(\dfrac{\sqrt{45mn^2}}{\sqrt{20m}}=\sqrt{\dfrac{45mn^2}{20m}}=\sqrt{\dfrac{9n^2}{4}}=\dfrac{3n}{2}\)

d) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{2\left|a\right|\sqrt{2}}=\dfrac{-1}{2a\sqrt{2}}\)