Sửa đề: Đưa thừa số vào trong dấu căn

a: \(3\sqrt{x^2}=\sqrt{3^2\cdot x^2}=\sqrt{9x^2}\)

b: \(-5\sqrt{y^4}=-\sqrt{5^2\cdot y^4}=-\sqrt{25y^4}\)

c: \(3\sqrt{5x}=\sqrt{3^2\cdot5x}=\sqrt{45x}\)

d: \(x\sqrt{7}=\sqrt{x^2\cdot7}=\sqrt{7x^2}\)

√(28a4b2 ) = √((2a2b)2.7) = √7 |2a2b| = 2√7a2b (do b ≥ 0)

a: \(\sqrt{5a^2}=\left|a\sqrt{5}\right|=-a\sqrt{5}\left(a< =0\right)\)

c: A=\(\sqrt{72a^2b^4}=\sqrt{36a^2b^4\cdot2}=6\sqrt{2}\cdot b^2\cdot\left|a\right|\)

mà a<0

nên \(A=-6\sqrt{2}\cdot ab^2\)

d: \(\sqrt{24a^4b^8}=\sqrt{4a^4b^8\cdot6}=2a^2b^4\cdot\sqrt{6}\)

b)  63 y 3 7 y với y > 0

=  63 y 3 7 y = 9 y 2 = 3 y = 3 y do y > 0

a) \(2\sqrt{5a^2}=2\sqrt{5}\left|a\right|=-2a\sqrt{5}\)

b)\(2\sqrt{18a^2}=2.3\sqrt{2}.\left|a\right|=6a\sqrt{2}\)

c)\(\sqrt{-9b^3}=\sqrt{9.\left(-b\right)^3}=3\sqrt{-b}.\left|b\right|=-3b\sqrt{-b}\)

d)\(\sqrt{24a^4b^8}=\sqrt{6.\left(4a^2b^4\right)^2}=2a^2b^4\sqrt{6}\)

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

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

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

25 x 3 = 25 x 2 . x = 5 x x = 5 x x ( với x > 0 )

7 x 2 = x 7 = x 7 ( với x > 0 )