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\(3\sqrt{5}=\sqrt{45}\)
\(-5\sqrt{2}=-\sqrt{25}.\sqrt{2}=-\sqrt{50}\)
\(\dfrac{-2}{3}\sqrt{xy}=-\sqrt{\dfrac{4}{9}}.\sqrt{xy}=-\sqrt{\dfrac{4}{9}xy}\left(xy\ge0\right)\)
\(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2}.\sqrt{\dfrac{2}{x}}=\sqrt{\dfrac{2x^2}{x}}=\sqrt{2x}\left(x>0\right)\)
Bài 2:
a: \(=\sqrt{\left(\dfrac{1}{5a}\right)^2}=\dfrac{1}{\left|5a\right|}=\dfrac{-1}{5a}\)
b: \(=\dfrac{1}{3}\cdot15\cdot\left|a\right|=5\left|a\right|\)
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)
b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)
c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
a) \(\sqrt{27x^2}=\sqrt{3.\left(3x\right)^2}=\left|3x\right|.\sqrt{3}=3x\sqrt{3}\left(x>0\right)\)
b) \(\sqrt{8xy^2}=\left|y\right|.2\sqrt{2x}=-2y\sqrt{2x}\left(x\ge0,y\le0\right)\)
1) \(x\sqrt{13}=\sqrt{13x^2}\left(x\ge0\right)\)
2) \(x\sqrt{-15x}=-\left|x\right|\sqrt{15x}=-\sqrt{15x^3}\left(x< 0\right)\)
3) \(x\sqrt{2}=-\left|x\right|\sqrt{2}=-\sqrt{2x^2}\left(x\le0\right)\)
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}\)
\(=\dfrac{1}{6}\sqrt{6}\)
b: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
a: \(=\sqrt{\left(2-a\right)^2\cdot\dfrac{2a}{a-2}}=\sqrt{2a\left(a-2\right)}\)
b: \(=\sqrt{\left(x-5\right)^2\cdot\dfrac{x}{\left(5-x\right)\left(5+x\right)}}\)
\(=\sqrt{\left(x-5\right)\cdot\dfrac{x}{x+5}}\)
c: \(=\sqrt{\left(a-b\right)^2\cdot\dfrac{3a}{\left(b-a\right)\left(b+a\right)}}=\sqrt{\dfrac{3a\left(b-a\right)}{b+a}}\)
a )\(\sqrt{5x^2}\)
b )\(-\sqrt{13x^2}\)
c )\(\sqrt{11x}\)
d)\(-\sqrt{-29x}\)
Bài 1: Đưa thừa số ra ngoài dấu căn:
\(2\sqrt{225a^2}=2.15a=30a\)
Bài 2: Đưa thừa số vào trong dấu căn :
\(x\sqrt{\dfrac{-39}{x}}=\sqrt{x^2.\dfrac{-39}{x}}=\sqrt{-39x}\)
Bài 3: Sắp xếp theo thứ tự tăng dần :
a) \(2\sqrt{3}< 3\sqrt{2}< 2\sqrt{5}< 5\sqrt{2}\)
b) \(4\sqrt{2}< \sqrt{37}< 2\sqrt{15}< 3\sqrt{7}\)
c) \(6\sqrt{\dfrac{1}{3}}< \sqrt{27}< 2\sqrt{28}< 5\sqrt{7}\)
3\(\sqrt{5}\)= \(\sqrt{3^2.5}\)=\(\sqrt{45}\)
-5\(\sqrt{2}\)= \(-\sqrt{5^2.2}\)= -\(\sqrt{50}\)
\(\dfrac{-2}{3}\sqrt{xy}\) = \(-\sqrt{\left(\dfrac{2}{3}\right)^2xy}\) = -\(\sqrt{\dfrac{4}{9}xy}\)
x\(\sqrt{\dfrac{2}{x}}\)= \(\sqrt{\dfrac{2x^2}{x}}=\sqrt{2x}\)
\(3\sqrt{5}=\sqrt{45}\)