Nếu sai mong bạn thông cảm:

\(\sqrt{50.6}=\sqrt{300}=\sqrt{2^2.5^2.3}=2.5\sqrt{3}=10\sqrt{3}\)

\(\sqrt{50\cdot6}=\sqrt{300}=10\sqrt{3}\)

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

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

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

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

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

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

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

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

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

\(\sqrt{34+24\sqrt{2}}=\sqrt{18+2.3\sqrt{2}.4+16}\)

\(=\sqrt{\left(3\sqrt{2}+4\right)^2}=3\sqrt{2}+4\)

 a/   \(\sqrt{a^4b^5}=a^2b^2\sqrt{b}\)

b/    \(\sqrt{a^6b^{11}}=a^3b^5\sqrt{b}\)

\(=3ab^2\)

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

\(\sqrt{\dfrac{45\left(x+y\right)^2.y^2}{16}}\left(dkxd:x\ge0,y\ge0\right)\\ =\dfrac{\sqrt{45\left(x+y\right)^2.y^2}}{\sqrt{16}}\\ =\dfrac{\sqrt{45}.\sqrt{\left(x+y\right)^2}.\sqrt{y^2}}{\sqrt{4^2}}\\ =\dfrac{\sqrt{3^2.5}.\left|x+y\right|.\left|y\right|}{4}\\ =\dfrac{3\sqrt{5}\left(x+y\right).y}{4}\\ =\dfrac{3\sqrt{5}\left(xy+y^2\right)}{4}\)

x^ 3 mà

y^3 nhầm