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\(\frac{1}{x-y}.\sqrt{x^4\left(x^2+y^2-2xy\right)}\)
\(=\frac{1}{x-y}.\sqrt{\left(x^2\right)^2.\left(x-y\right)^2}\)
\(=\frac{1}{x-y}\left(x-y\right)x^2\)
\(=x^2\)
\(\frac{2xy^2}{3ab}\sqrt{\frac{9a^3b^4}{8xy^3}}=\frac{2xy^2}{3ab}\frac{3\sqrt{a^2.a}\sqrt{\left(b^2\right)^2}}{2\sqrt{2xy^2.y}}\)
\(=\frac{2xy^2}{3ab}\frac{3a\sqrt{a}b^2}{2y\sqrt{2xy}}=\frac{6xy^2ab^2\sqrt{a}}{6aby\sqrt{2xy}}=\frac{bxy\sqrt{a}}{\sqrt{2xy}}\)
\(=\frac{bxy\sqrt{2axy}}{2xy}=\frac{b\sqrt{2axy}}{2}\)
\(\sqrt{\left(-9\right)\cdot\left(-36\right)\cdot ab^2}\)
\(=\sqrt{9\cdot36\cdot ab^2}\)
\(=3\cdot4\cdot\left|b\right|\cdot\sqrt{a}\)
\(=12\left|b\right|\cdot\sqrt{a}\)
b) 63 y 3 7 y với y > 0
= 63 y 3 7 y = 9 y 2 = 3 y = 3 y do y > 0
a: \(a^2\cdot\sqrt{\dfrac{2}{3a}}=a^2\cdot\dfrac{\sqrt{2}}{\sqrt{3}\cdot\sqrt{a}}=\dfrac{a\sqrt{2}}{\sqrt{3}}=\dfrac{a\sqrt{6}}{3}\)
b: \(\dfrac{x-3}{x}\cdot\sqrt{\dfrac{x^3}{9-x^2}}\)
\(=\dfrac{x-3}{x}\cdot\dfrac{x\sqrt{x}}{\sqrt{x-3}\cdot\sqrt{x+3}}\)
\(=\dfrac{\sqrt{x}\cdot\sqrt{x-3}}{\sqrt{x+3}}\)
