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a,\(-\sqrt{10x^2\cdot y\left(3-\sqrt{2}\right)^2}=-\left|x\right|\) \(\cdot\left(3-\sqrt{2}\right)\cdot\sqrt{10y}\)
xet th \(x\ge0\) ta co \(-x\cdot\left(3-\sqrt{2}\right)\sqrt{10y}\)
xet th \(x< 0\) ta có \(x\left(3-\sqrt{2}\right)\sqrt{10y}\)
b,\(\sqrt{3\left(x^2-2xy+y^2\right)}=\) \(\sqrt{3\cdot\left(x-y\right)^2}=\left|x-y\right|\sqrt{3}\)
\(\frac{2xy^2}{3ab}\sqrt{\frac{9a^3b^4}{8xy^3}}=\frac{2xy^2}{3ab}\frac{3\sqrt{a^2.a}\sqrt{\left(b^2\right)^2}}{2\sqrt{2xy^2.y}}\)
\(=\frac{2xy^2}{3ab}\frac{3a\sqrt{a}b^2}{2y\sqrt{2xy}}=\frac{6xy^2ab^2\sqrt{a}}{6aby\sqrt{2xy}}=\frac{bxy\sqrt{a}}{\sqrt{2xy}}\)
\(=\frac{bxy\sqrt{2axy}}{2xy}=\frac{b\sqrt{2axy}}{2}\)
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) \(\sqrt{27x^2}=\sqrt{3.\left(3x\right)^2}=\left|3x\right|.\sqrt{3}=3x\sqrt{3}\left(x>0\right)\)
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) \(\sqrt{25\cdot96}=\sqrt{5^2\cdot2^5\cdot3}=\sqrt{5^2\cdot\left(2^2\right)^2\cdot2\cdot3}\)
\(=20\sqrt{6}\)
b) \(\sqrt{21\cdot75\cdot14}=\sqrt{2\cdot3^2\cdot5^2\cdot7^2}=105\sqrt{2}\)
c) \(y^2\sqrt{x^6\cdot y^8}=\sqrt{x^6\cdot y^4\cdot y^8}=\sqrt{\left(x^3\right)^2\cdot\left(y^6\right)^2}=x^3\cdot y^6\)
hì,giúp bn đc phần a thôi nha!!!
\(a,\sqrt{25.96}=\sqrt{25.16.6}=\sqrt{25}.\sqrt{16}.\sqrt{6}=5.4.\sqrt{6}=20\sqrt{6}\)
=.= hok tốt!!!
Giải:
a) \(\sqrt{a^4.\left(3-a\right)^2}\)
\(=\sqrt{\left(a^2\left(3-a\right)\right)^2}\)
\(=\left|a^2\left(3-a\right)\right|\)
b) \(\sqrt{27.48.\left(a-3\right)^2}\)
\(=\sqrt{3.9.16.3.\left(a-3\right)^2}\)
\(=\sqrt{3.3.9.16\left(a-3\right)^2}\)
\(=\sqrt{\left(9.4\left(a-3\right)\right)^2}\)
\(=\left|9.4\left(a-3\right)\right|\)
\(=\left|36\left(a-3\right)\right|\)
c) \(\sqrt{48.75a^2}\)
\(=\sqrt{16.3.25.3a^2}\)
\(=\sqrt{\left(4.3.5a\right)^2}\)
\(=\left|4.3.5a\right|\)
\(=\left|60a\right|\)
d) \(\sqrt{2^4.\left(-9\right)^2}\)
\(=\sqrt{2^4.9^2}\)
\(=\sqrt{\left(2^2.9\right)^2}\)
\(=\left|2^2.9\right|\)
\(=\left|36\right|=36\)
Vậy ...
Bài 1:
\(\sqrt{27a^2}=3a\sqrt{3}\)
Bài 2:
\(\dfrac{2}{3}\sqrt{3xy}=\sqrt{3xy\cdot\dfrac{4}{9}}=\sqrt{\dfrac{4}{3}xy}\)
Bài 3:
\(=4\sqrt{b}+2\cdot2\sqrt{10b}-3\cdot3\sqrt{10b}=4\sqrt{b}-5\sqrt{10b}\)
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 \(\frac{3}{4}.\sqrt{7}>\frac{2}{5}.\sqrt{5}\)
