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

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

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

\(\frac{1}{x-y}.\sqrt{x^4\left(x^2+y^2-2xy\right)}\)

\(=\frac{1}{x-y}.\sqrt{\left(x^2\right)^2.\left(x-y\right)^2}\)

\(=\frac{1}{x-y}\left(x-y\right)x^2\)

\(=x^2\)

chịu.-.

HT~~~

a, -2x^2y căn 7 

b, ab^2 căn 63 

c, a-1 căn 147a-147

d, y+2 nhân căn [192 nhân (y+2)^3]

a)-2x²y√7

b) 3ab²√7

c) 7(a-1)√3(a-1)

d) 8(y+2)²√3(y+2)

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

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

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)`.

đéo biết làm, đăng làm quái gì không biết -_-"

a) \(\sqrt{\frac{9a^2-12ab+4b^2}{81a^4b^4}}=\sqrt{\frac{\left(3a-4b\right)^2}{\left(9a^2b^2\right)^2}}\)

\(=\frac{3a-4b}{9a^2b^2}\)

b)\(\sqrt{\frac{1}{a}-\frac{1}{a^2}}=\sqrt{\frac{a-1}{a^2}}=\frac{1}{a}\sqrt{a-1}\)

P/s tham khảo nhé

đưa thừa số vào trong dấu căn*

mình nhầm

a: \(a^2\cdot\sqrt{\dfrac{2}{3a}}=a^2\cdot\dfrac{\sqrt{2}}{\sqrt{3}\cdot\sqrt{a}}=\dfrac{a\sqrt{2}}{\sqrt{3}}=\dfrac{a\sqrt{6}}{3}\)

b: \(\dfrac{x-3}{x}\cdot\sqrt{\dfrac{x^3}{9-x^2}}\)

\(=\dfrac{x-3}{x}\cdot\dfrac{x\sqrt{x}}{\sqrt{x-3}\cdot\sqrt{x+3}}\)

\(=\dfrac{\sqrt{x}\cdot\sqrt{x-3}}{\sqrt{x+3}}\)