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)`.

Nếu sai mong bạn thông cảm:

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

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

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

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

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

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

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

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

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

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

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

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

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

a/ \(\sqrt{50a}=5\sqrt{2a}\)

b/ \(\sqrt{75x}=5\sqrt{3x}\)

a)\(5\sqrt{2a}\)

b)\(5\sqrt{3x}\)

a) \(\sqrt{50a}=\sqrt{25.2}=5\sqrt{2a}\)

b) \(\sqrt{75x}=\sqrt{25.3x}=5\sqrt{3x}\)

\(\sqrt{6-4\sqrt{2}}=\sqrt{\left(2^2\right)-2.2.\sqrt{2}+\left(\sqrt{2}\right)^2}\)

\(=\sqrt{\left(2-\sqrt{2}\right)^2}\)=/\(2-\sqrt{2}\)/=\(2-\sqrt{2}\)(vì 2>\(\sqrt{2}\))

        \(\sqrt{6-4\sqrt{2}}\)

\(\Leftrightarrow\sqrt{2^2-2.2.\sqrt{2}+\left(\sqrt{2}\right)^2}\)

\(\Leftrightarrow\sqrt{\left(2-\sqrt{2}\right)^2}\)

\(\Leftrightarrow2-\sqrt{2}\)