Bài 1.
a) \(\sqrt{50}=\sqrt{25}.\sqrt{2}=5\sqrt{2}\)
b) \(\sqrt{72}=\sqrt{36}.\sqrt{2}=6\sqrt{2}\)
c) \(\sqrt{180x^2}=\sqrt{36x^2}.\sqrt{5}=6\left|x\right|\sqrt{5}\)
d) \(\sqrt{4x^2y^4}=\sqrt{4}.\sqrt{x^2}.\sqrt{y^4}=2\left|x\right|y^2\)
Bài 2.
a) \(2\sqrt{24}=\sqrt{2^2.24}=\sqrt{96}\)
b) \(-3\sqrt{15}=-\sqrt{3^2.15}=-\sqrt{135}\)
c) \(a\sqrt{\dfrac{3}{a}}=\sqrt{\dfrac{a^2.3}{a}}=\sqrt{3a}\)
d) \(a\sqrt{2}=\sqrt{2a^2}\left(a\ge0\right)\)
Hoặc \(a\sqrt{2}=-\sqrt{2a^2}\left(a< 0\right)\)
Bài 3.
a) \(3\sqrt{11}=\sqrt{3^2.11}=\sqrt{99}\)
\(2\sqrt{15}=\sqrt{2^2.15}=\sqrt{60}\)
Do \(99>60\Rightarrow\sqrt{99}>\sqrt{60}\)
Vậy \(3\sqrt{11}>2\sqrt{15}\)
b) \(\dfrac{1}{4}\sqrt{5}=\sqrt{\dfrac{1}{16}.5}=\sqrt{\dfrac{5}{16}}\)
\(5\sqrt{\dfrac{1}{4}}=\sqrt{\dfrac{25.1}{4}}=\sqrt{\dfrac{25}{4}}\)
Do \(\dfrac{25}{4}>\dfrac{4}{4}=1;1=\dfrac{16}{16}>\dfrac{5}{16}\Rightarrow\dfrac{25}{4}>\dfrac{5}{16}\)\(\Rightarrow\sqrt{\dfrac{25}{4}}>\sqrt{\dfrac{5}{16}}\)
Vậy \(5\sqrt{\dfrac{1}{4}}>\dfrac{1}{4}\sqrt{5}\)
Bài 4.
a) \(\sqrt{72}-3\sqrt{20}-5\sqrt{2}+\sqrt{180}\)
\(=6\sqrt{2}-6\sqrt{5}-5\sqrt{2}+6\sqrt{5}\)
\(=\sqrt{2}\)
b) \(2\sqrt{3x}-\sqrt{48x}+\sqrt{108x}+\sqrt{3x}\)
\(=2\sqrt{3x}-4\sqrt{3x}+6\sqrt{3x}+\sqrt{3x}\)
\(=5\sqrt{3x}\)
Bài 5.
\(\dfrac{1}{1-5x}.\sqrt{3x^2\left(25x^2-10x+1\right)}\)
\(=\dfrac{1}{1-5x}.\sqrt{3x^2.\left(1-5x\right)^2}\)
\(=\dfrac{x.\left(1-5x\right).\sqrt{3}}{1-5x}\)
\(=\sqrt{3}x\)
Bài 1
`sqrt{50} = sqrt{5^2 . 2} = 5sqrt{2}`
`sqrt{72} = sqrt{6^2 . 2} = 6sqrt{2}`
`sqrt{180x^2} = sqrt{5.(6x)^2} = 6xsqrt{5}`
`sqrt{4x^2y^4} = sqrt{(2xy^2)^2} = |2xy^2 | = 2xy^2`
