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d) \(\dfrac{1}{3}\sqrt{225a^2}=\dfrac{1}{3}\sqrt{\left(15a\right)^2}=\dfrac{1}{3}\left|15a\right|=\left|5a\right|\)
\(\Rightarrow\left[{}\begin{matrix}a>0\Rightarrow d=5a\\a< 0\Rightarrow d=-5a\end{matrix}\right.\)
Giải:
a) \(\sqrt{49.360}\)
\(=\sqrt{7^2.3^2.2^2.10}\)
\(=7.3.2\sqrt{10}\)
\(=42\sqrt{10}\)
Vậy ...
b) \(-\sqrt{500.162}\)
\(=-\sqrt{10^2.5.9^2.2}\)
\(=-10.9\sqrt{10}\)
\(=-90\sqrt{10}\)
Vậy ...
c) \(\sqrt{125a^2}\)
\(=\sqrt{5^2.5.a^2}\)
\(=\sqrt{5^2.5.\left(-a\right)^2}\)
\(=-5a\sqrt{5}\)
Vậy ...
d) \(\dfrac{1}{3}\sqrt{225.a^2}\)
\(=\dfrac{1}{3}\sqrt{15^2.a^2}\)
\(=\dfrac{1}{3}.15.a^2\)
\(=5a^2\)
Vậy ...
Lời giải:
\(\sqrt{\frac{(1+\sqrt{2})^3}{27}}=\sqrt{\frac{(1+\sqrt{2})^3}{3^3}}=\sqrt{\frac{3(1+\sqrt{2})^3}{3^4}}\)
\(=\frac{(1+\sqrt{2})\sqrt{3+3\sqrt{2}}}{9}\)
\(ab\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{(ab)^2(\frac{1}{a}+\frac{1}{b})}=\sqrt{ab^2+a^2b}\)
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) \(6\sqrt{\dfrac{1}{3}}< \sqrt{27}< 2\sqrt{28}< 5\sqrt{7}\)
a) \(\sqrt{49.360}=\sqrt{7^2.6^2.10}=7.6\sqrt{10}=42\sqrt{10}\)
b)\(\sqrt{125a^2}=\sqrt{5^2.5.a^2}=5.\left|a\right|\sqrt{5}=-5a\sqrt{5}\) ( vì a<0)
c)\(-\sqrt{500.162}=-\sqrt{10^2.5.9^2.2}=-10.9\sqrt{5.2}=-90\sqrt{10}\)
d) \(\frac{1}{3}\sqrt{225a^2}=\frac{1}{3}\sqrt{15^2.a^2}=\frac{1}{3}.15.\left|a\right|=\frac{15a}{3}\) ( a>0)
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)Ta có: \(2\sqrt{5}< 5\sqrt{2}\)\(2\sqrt{5}=\sqrt{2^2.5}=\sqrt{20}\)
\(5\sqrt{2}=\sqrt{5^2.2}=\sqrt{50}\)
Vì \(\sqrt{20}< \sqrt{50}\)
Nên \(2\sqrt{5}< 5\sqrt{2}\)
b)Ta có: \(3\sqrt{13}=\sqrt{3^2.13}=\sqrt{117}\)
\(4\sqrt{11}=\sqrt{4^2.11}=\sqrt{176}\)
Vì \(\sqrt{117}< \sqrt{176}\)
Nên \(3\sqrt{13}< 4\sqrt{11}\)
c) Ta có: \(\frac{3}{4}.\sqrt{7}=\sqrt{\left(\frac{3}{4}\right)^2.7}=\sqrt{\frac{63}{16}}\)
\(\frac{2}{5}.\sqrt{5}=\sqrt{\left(\frac{2}{5}\right)^2.5}=\sqrt{\frac{4}{5}}\)
Vì \(\sqrt{\frac{63}{16}}>1\)
\(\sqrt{\frac{4}{5}}< 1\)
Nên \(\sqrt{\frac{63}{16}}>\sqrt{\frac{4}{5}}\)
Vậy \(\frac{3}{4}.\sqrt{7}>\frac{2}{5}.\sqrt{5}\)
a )\(x\sqrt{7}\)
b )\(-2y\sqrt{2}\)
c )\(5x\sqrt{x}\)
d)\(4y^2\sqrt{3}\)
a) \(\sqrt{49.360}=\sqrt{7^2.6^2.10}=7.6\sqrt{10}=42\sqrt{10}\)
b) \(-\sqrt{500.162}=-\sqrt{\left(2^2.5^2.5\right).\left(2.3^2.3^2\right)}\)
\(=-2.5.3.3\sqrt{5.2}=-90\sqrt{10}\)
c) \(\sqrt{125a^2}=\sqrt{5^2.a^2.5}=5\left|a\right|\sqrt{5}=-5a\sqrt{5}\) (do a <0)
d) \(\frac{1}{3}\sqrt{225a^2}=\frac{1}{3}\sqrt{15^2.a^2}=\frac{15}{3}\left|a\right|=5\left|a\right|\)
a < 0 thì suy ra \(\frac{1}{3}\sqrt{225a^2}=\frac{1}{3}\sqrt{15^2.a^2}=\frac{15}{3}\left|a\right|=5\left|a\right|=-5a\)
a>=0 thì suy ra \(\frac{1}{3}\sqrt{225a^2}=\frac{1}{3}\sqrt{15^2.a^2}=\frac{15}{3}\left|a\right|=5\left|a\right|=5a\)