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a: \(=3\sqrt{3}\cdot x\)

b: \(=-2\sqrt{2x}\cdot y\sqrt{y}\)

c: \(=5x\sqrt{x}\)

d: \(=\sqrt{42x}\cdot y^2\)

9 tháng 10 2021

\(a,=6\left|a\right|b^2\sqrt{2}=6ab^2\sqrt{2}\\ b,=3\left|ab\right|\sqrt{3a}=-3ab\sqrt{3a}\)

9 tháng 10 2021

\(a,=\sqrt{12a^4b}\\ b,\sqrt{18\left(-a\right)^4b^8}\)

17 tháng 8 2020

a) \(\sqrt{27x^2}=\sqrt{3.\left(3x\right)^2}=\left|3x\right|.\sqrt{3}=3x\sqrt{3}\left(x>0\right)\)

b) \(\sqrt{8xy^2}=\left|y\right|.2\sqrt{2x}=-2y\sqrt{2x}\left(x\ge0,y\le0\right)\)

1) \(x\sqrt{13}=\sqrt{13x^2}\left(x\ge0\right)\)

2) \(x\sqrt{-15x}=-\left|x\right|\sqrt{15x}=-\sqrt{15x^3}\left(x< 0\right)\)

3) \(x\sqrt{2}=-\left|x\right|\sqrt{2}=-\sqrt{2x^2}\left(x\le0\right)\)

25 tháng 7 2021

\(\frac{2xy^2}{3ab}\sqrt{\frac{9a^3b^4}{8xy^3}}=\frac{2xy^2}{3ab}\frac{3\sqrt{a^2.a}\sqrt{\left(b^2\right)^2}}{2\sqrt{2xy^2.y}}\)

\(=\frac{2xy^2}{3ab}\frac{3a\sqrt{a}b^2}{2y\sqrt{2xy}}=\frac{6xy^2ab^2\sqrt{a}}{6aby\sqrt{2xy}}=\frac{bxy\sqrt{a}}{\sqrt{2xy}}\)

\(=\frac{bxy\sqrt{2axy}}{2xy}=\frac{b\sqrt{2axy}}{2}\)

18 tháng 1 2022

chịu.-.

HT~~~

3 tháng 10 2022

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19 tháng 10 2021

\(\sqrt{48\cdot45}=12\sqrt{15}\\ \sqrt{225\cdot17}=15\sqrt{17}\\ \sqrt{a^3b^7}=\left|ab^3\right|\sqrt{ab}=ab^3\sqrt{ab}\\ \sqrt{x^5\left(x-3\right)^2}=\left|x^2\left(x-3\right)\right|\sqrt{x}=x^2\left(x-3\right)\sqrt{x}\)

19 tháng 10 2021

\(\sqrt{48\cdot45}=4\sqrt{3}\cdot3\sqrt{5}=12\sqrt{15}\)

\(\sqrt{225\cdot17}=15\sqrt{17}\)

21 tháng 7 2023

a) \(\sqrt{27x^2}\)

\(=\sqrt{3^2\cdot3x^2}\)

\(=\left|3x\right|\sqrt{3}\)

\(=3\left|x\right|\sqrt{3}\)

b) \(\sqrt{8xy^2}\)

\(=\sqrt{2^2\cdot2\cdot x\cdot y^2}\)

\(=\left|2y\right|\sqrt{2x}\)

\(=2\left|y\right|\sqrt{2x}\)

c) \(\sqrt{25x^3}\)

\(=\sqrt{5^2\cdot x^2\cdot x}\)

\(=\left|5x\right|\sqrt{x}\)

\(=5\left|x\right|\sqrt{x}\)

d) \(\sqrt{48xy^4}\)

\(=\sqrt{4^2\cdot3x\cdot\left(y^2\right)^2}\)

\(=\left|4y^2\right|\sqrt{3x}\)

\(=4y^2\sqrt{3x}\)

`a, sqrt(27x^2b) = sqrt(3^2. 3.x^2b) = 3|x|sqrt(3b)`.

`b, sqrt(8xy^2) =sqrt(2^2.2xy^2)= 2|y|sqrt(2x)`

`c, sqrt(25x^3d) = sqrt(5^2.x^2.x.d) = 5|x|sqrt(xd)`.

`d, sqrt(48xy^4) = sqrt(4^2.3 . xy^4) = 4y^2sqrt(3x)`.

1 tháng 10 2023

a) \(\dfrac{y}{x}\cdot\sqrt{\dfrac{x^2}{y^4}}\)

\(=\dfrac{y}{x}\cdot\dfrac{\sqrt{x^2}}{\sqrt{\left(y^2\right)^2}}\) 

\(=\dfrac{y}{x}\cdot\dfrac{x}{y^2}\)

\(=\dfrac{1}{y}\)

b) \(\dfrac{5}{2}x^3y^3\cdot\sqrt{\dfrac{16}{x^4y^8}}\)

\(=\dfrac{5}{2}x^3y^3\cdot\dfrac{\sqrt{16}}{\sqrt{\left(x^2y^4\right)^2}}\)

\(=\dfrac{5}{2}x^3y^3\cdot\dfrac{4}{x^2y^4}\)

\(=\dfrac{20x^3y^3}{2x^2y^4}\)

\(=\dfrac{10x}{y}\)

c) \(ab^2\sqrt{\dfrac{3}{a^2b^4}}\)

\(=ab^2\dfrac{\sqrt{3}}{\sqrt{\left(ab^2\right)^2}}\)

\(=ab^2\cdot\dfrac{\sqrt{3}}{ab^2}\)

\(=\sqrt{3}\)

1 tháng 10 2023

\(a,\dfrac{y}{x}\cdot\sqrt{\dfrac{x^2}{y^4}}\left(y\ge0;x,y\ne0\right)\) (sửa đề)

\(=\dfrac{y}{x}\cdot\dfrac{\sqrt{x^2}}{\sqrt{y^4}}\)

\(=\dfrac{y}{x}\cdot\dfrac{x}{\sqrt{\left(y^2\right)^2}}\)

\(=\dfrac{y}{x}\cdot\dfrac{x}{y^2}\)

\(=\dfrac{1}{y}\)

\(---\)

\(b,\dfrac{5}{2}x^3y^3\cdot\sqrt{\dfrac{16}{x^4y^8}}\left(x,y\ne0\right)\)

\(=\dfrac{5}{2}x^3y^3\cdot\dfrac{\sqrt{16}}{\sqrt{x^4y^8}}\)

\(=\dfrac{5x^3y^3}{2}\cdot\dfrac{4}{x^2y^4}\)

\(=\dfrac{5x\cdot2}{y}\)

\(=\dfrac{10x}{y}\)

\(---\)

\(c,ab^2\sqrt{\dfrac{3}{a^2b^4}}\left(a>0;b\ne0\right)\) (sửa đề)

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

\(=\dfrac{ab^2\sqrt{3}}{\sqrt{\left(ab^2\right)^2}}\)

\(=\dfrac{ab^2\sqrt{3}}{ab^2}\)

\(=\sqrt{3}\)

#\(Toru\)

14 tháng 6 2017

a )\(x\sqrt{7}\)

b )\(-2y\sqrt{2}\)

c )\(5x\sqrt{x}\)

d)\(4y^2\sqrt{3}\)

1 tháng 10 2023

a) \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}\)

\(=\sqrt{\dfrac{2a\cdot3a}{3\cdot8}}\)

\(=\sqrt{\dfrac{6a^2}{24}}\)

\(=\sqrt{\dfrac{a^2}{4}}\)

\(=\dfrac{\sqrt{a^2}}{\sqrt{4}}\)

\(=\dfrac{a}{2}\)

b) \(\sqrt{3a}\cdot\sqrt{\dfrac{52}{a}}\)

\(=\sqrt{3a\cdot\dfrac{52}{a}}\) 

\(=\sqrt{3\cdot52}\)

\(=\sqrt{13\cdot3\cdot4}\)

\(=2\sqrt{39}\)

c) \(2y^2\cdot\sqrt{\dfrac{x^4}{4y^2}}\)

\(=2y^2\cdot\dfrac{\sqrt{\left(x^2\right)^2}}{\sqrt{\left(2y\right)^2}}\)

\(=2y^2\cdot\dfrac{x^2}{-2y}\)

\(=\dfrac{2y^2\cdot x^2}{-2y}\)

\(=-x^2y\)