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

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

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

ĐỀ bài bài 2 là: Đưa thừa số vào dấu căn bậc hai

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

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

\(\left(x-5\right)\cdot\sqrt{\dfrac{x}{25-x^2}}=\sqrt{\left(x-5\right)^2\cdot\dfrac{x}{-\left(x-5\right)\left(x+5\right)}}=\sqrt{-\dfrac{x\left(x-5\right)}{x+5}}\)

\(x\sqrt{\dfrac{7}{x^2}}=\sqrt{x^2\cdot\dfrac{7}{x^2}}=\sqrt{7}\)

\(\sqrt{18b^3\cdot\left(1-2a\right)^2}\)

\(=3\sqrt{2}\cdot b\sqrt{b}\cdot\left|1-2a\right|\)

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

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

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

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

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

3\(\sqrt{5}\)= \(\sqrt{3^2.5}\)=\(\sqrt{45}\)

-5\(\sqrt{2}\)= \(-\sqrt{5^2.2}\)= -\(\sqrt{50}\)

\(\dfrac{-2}{3}\sqrt{xy}\) = \(-\sqrt{\left(\dfrac{2}{3}\right)^2xy}\) = -\(\sqrt{\dfrac{4}{9}xy}\)

x\(\sqrt{\dfrac{2}{x}}\)= \(\sqrt{\dfrac{2x^2}{x}}=\sqrt{2x}\)

\(3\sqrt{5}=\sqrt{45}\)

\(-5\sqrt{2}=-\sqrt{50}\)