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\(\frac{2xy^2}{3ab}\sqrt{\frac{9a^3b^4}{8xy^3}}=\frac{2xy^2}{3ab}\frac{3\sqrt{a^2.a}\sqrt{\left(b^2\right)^2}}{2\sqrt{2xy^2.y}}\)
\(=\frac{2xy^2}{3ab}\frac{3a\sqrt{a}b^2}{2y\sqrt{2xy}}=\frac{6xy^2ab^2\sqrt{a}}{6aby\sqrt{2xy}}=\frac{bxy\sqrt{a}}{\sqrt{2xy}}\)
\(=\frac{bxy\sqrt{2axy}}{2xy}=\frac{b\sqrt{2axy}}{2}\)
a, Để A nhận giá trị dương thì \(A>0\)hay \(x-1>0\Leftrightarrow x>1\)
b, \(B=2\sqrt{2^2.5}-3\sqrt{3^2.5}+4\sqrt{4^2.5}\)
\(=4\sqrt{5}-9\sqrt{5}+16\sqrt{5}=\left(4-9+16\right)\sqrt{5}=11\sqrt{5}\)
( theo công thức \(A\sqrt{B}=\sqrt{A^2B}\))
c, Với \(a\ge0;a\ne1\)
\(C=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2.\frac{1}{\left(\sqrt{a}+1\right)^2}=1\)
a )\(x\sqrt{7}\)
b )\(-2y\sqrt{2}\)
c )\(5x\sqrt{x}\)
d)\(4y^2\sqrt{3}\)
\(\sqrt{48\cdot45}=12\sqrt{15}\\ \sqrt{225\cdot17}=15\sqrt{17}\\ \sqrt{a^3b^7}=\left|ab^3\right|\sqrt{ab}=ab^3\sqrt{ab}\\ \sqrt{x^5\left(x-3\right)^2}=\left|x^2\left(x-3\right)\right|\sqrt{x}=x^2\left(x-3\right)\sqrt{x}\)
\(\sqrt{48\cdot45}=4\sqrt{3}\cdot3\sqrt{5}=12\sqrt{15}\)
\(\sqrt{225\cdot17}=15\sqrt{17}\)
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) \(\sqrt{49.360}=\sqrt{7^2.6^2.10}=7.6\sqrt{10}=42\sqrt{10}\)
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 \(\frac{1}{3}\sqrt{225a^2}=\frac{1}{3}\sqrt{15^2.a^2}=\frac{15}{3}\left|a\right|=5\left|a\right|=5a\)
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 ...
chịu.-.
HT~~~