Rút gọn:
a,\(\sqrt{4\left(a-3\right)^2}với\) \(a\ge3\)
\(b,\sqrt{9\left(b-2\right)^2}với\) \(b< 2\)
\(c,\sqrt{27.48\left(1-a\right)^2}với\) \(a>1\)
\(d,\sqrt{5a}.\sqrt{45a}-3a\) \(với\) \(a\ge0\)
\(e,\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}với\)...
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Rút gọn:
a,\(\sqrt{4\left(a-3\right)^2}với\) \(a\ge3\)
\(b,\sqrt{9\left(b-2\right)^2}với\) \(b< 2\)
\(c,\sqrt{27.48\left(1-a\right)^2}với\) \(a>1\)
\(d,\sqrt{5a}.\sqrt{45a}-3a\) \(với\) \(a\ge0\)
\(e,\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}với\) \(x>0\)
a)\(\sqrt{4\left(a-3\right)^2}=\sqrt{2^2\left(a-3\right)^2}=\sqrt{\left(2a-6\right)^2}=2a-6\)
b) \(\sqrt{9\left(b-2\right)^2}=\sqrt{3^2\left(b-2\right)^2}=\sqrt{\left[3\left(b-2\right)\right]^2}=3b-6\)
c) bạn xem lại đề
d)
\(\sqrt{5a}.\sqrt{45a}-3a=\sqrt{225a^2}-3a=\sqrt{\left(15a\right)^2}-3a=15a-3a=12a\)
e) \(\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\dfrac{48x^3}{3x^5}}=\sqrt{\dfrac{16}{x^2}}=\dfrac{\sqrt{16}}{\sqrt{x^2}}=\dfrac{4}{x}\)