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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)\(\sqrt{24a^4b^8}=\sqrt{6.\left(4a^2b^4\right)^2}=2a^2b^4\sqrt{6}\)
a: \(\sqrt{36\cdot3\cdot\left(a+7\right)^2}=6\sqrt{3}\left|a+7\right|\)
b: \(\sqrt{9^2\cdot a^4\cdot b^3\cdot b^3\cdot b}=9a^2b^3\sqrt{b}\)
c: Nếu đk xác định như này thì \(C=\sqrt{16a^5b^3}\) chỉ xác định với a=b=0 thôi nha bạn
=>C=0
a: \(\sqrt{48a^4b^2}=\sqrt{16a^4b^2\cdot3}=4\sqrt{3}\cdot a^2\cdot\left|b\right|\)
\(=-4\sqrt{3}\cdot a^2b\)
b: \(\sqrt{-25x^3}=\sqrt{-25x^2\cdot x}=\left|25x^2\right|\cdot\sqrt{-x}\)
\(=-5x\sqrt{-x}\)
\(\sqrt{18b^3\cdot\left(1-2a\right)^2}\)
\(=3\sqrt{2}\cdot b\sqrt{b}\cdot\left|1-2a\right|\)
\(=3\sqrt{2}\left(2a-1\right)\cdot b\sqrt{b}\)
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: \(\dfrac{x-3}{x}\cdot\sqrt{\dfrac{x^3}{9-x^2}}\)
\(=\dfrac{x-3}{x}\cdot\dfrac{x\sqrt{x}}{\sqrt{x-3}\cdot\sqrt{x+3}}\)
\(=\dfrac{\sqrt{x}\cdot\sqrt{x-3}}{\sqrt{x+3}}\)
\(a,=6\left|a\right|b^2\sqrt{2}=6ab^2\sqrt{2}\\ b,=3\left|ab\right|\sqrt{3a}=-3ab\sqrt{3a}\)
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: \(\sqrt{24a^4b^8}=\sqrt{4a^4b^8\cdot6}=2a^2b^4\cdot\sqrt{6}\)
1) \(ab^4\sqrt{a}=\sqrt{\left(ab^4\right)^2a}=\sqrt{a^2b^8a}=\sqrt{a^3b^8}\)
2) \(-2ab^2\sqrt{5a}=-\sqrt{\left(-2ab^2\right)^25a}=\sqrt{4a^2b^45a}\)
\(\sqrt{20a^3b^4}\)