Đưa thừa số ra ngoài dấu căn :
a) \(4\sqrt{3\:}\) + \(\sqrt{27}\) - \(\sqrt{45}\) + \(\sqrt{5}\)
b) \(\sqrt{28a^4b^{2\:}}\) với b ≥ 0
c) \(\sqrt{72a^2b^4}\) với a < 0
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a: \(=\sqrt{4\cdot a^4b^2\cdot7}=2a^2b\sqrt{7}\left(b>=0\right)\)
b: \(=\sqrt{36\cdot b^4\cdot a^2\cdot2}=-6ab^2\sqrt{2}\)
\(a,=6\left|a\right|b^2\sqrt{2}=6ab^2\sqrt{2}\\ b,=3\left|ab\right|\sqrt{3a}=-3ab\sqrt{3a}\)
a/ \(0,1\sqrt{2.10000=0,1\sqrt{ }2.100^{ }2=0,1\cdot100\sqrt{ }2=10\sqrt{ }2}\)
b/ \(-0,05\sqrt{28800}=-0,05\sqrt{288\cdot100=-0,05\cdot10\sqrt{ }288=6\sqrt{ }2}\)
c/\(\sqrt{7\cdot63}a^2=\sqrt{7\cdot9\cdot7}a^2=21a^2\)
\(\sqrt{72a^{ }2b\sqrt{ }4=\sqrt{ }6\cdot9\left|\right|ab^{ }2=-3\sqrt{ }6ab^{ }2}\)
a: ĐKXĐ: 2x-10>=0
=>2x>=10
=>x>=5
b: \(\sqrt{A^2B}=\sqrt{A^2}\cdot\sqrt{B}=\left|A\right|\cdot\sqrt{B}\)
\(\sqrt{72}=\sqrt{36\cdot2}=6\sqrt{2}\)
c: \(A=\sqrt{16}+\sqrt{81}=4+9=13\)
\(B=\sqrt{\dfrac{\left(15\sqrt{5}+5\sqrt{200}-3\sqrt{450}\right)}{\sqrt{10}}}\)
\(=\sqrt{\dfrac{15}{\sqrt{2}}+5\sqrt{20}-3\sqrt{45}}\)
\(=\sqrt{\dfrac{15\sqrt{2}+2\sqrt{5}}{2}}=\sqrt{\dfrac{30\sqrt{2}+4\sqrt{5}}{4}}\)
\(=\dfrac{\sqrt{30\sqrt{2}+4\sqrt{5}}}{2}\)
\(C=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\left(2+\sqrt{3}\right)\)
\(=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}-\left(2+\sqrt{3}\right)+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
\(=2+\sqrt{3}-2-\sqrt{3}+\sqrt{2}=\sqrt{2}\)
\(\sqrt{48\cdot45}=12\sqrt{15}\\ \sqrt{225\cdot17}=15\sqrt{17}\\ \sqrt{a^3b^7}=\left|ab^3\right|\sqrt{ab}=ab^3\sqrt{ab}\\ \sqrt{x^5\left(x-3\right)^2}=\left|x^2\left(x-3\right)\right|\sqrt{x}=x^2\left(x-3\right)\sqrt{x}\)
\(\sqrt{48\cdot45}=4\sqrt{3}\cdot3\sqrt{5}=12\sqrt{15}\)
\(\sqrt{225\cdot17}=15\sqrt{17}\)
a) = \(4\sqrt{3}+3\sqrt{3}-3\sqrt{5}+\sqrt{5}=\sqrt{3}\cdot\left(4+3\right)-\sqrt{5}\cdot\left(3-1\right)=7\sqrt{3}-2\sqrt{5}\)
b) = \(2a^2b\sqrt{7b}\)
c) = \(6ab^2\sqrt{2}\)