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a )\(x\sqrt{7}\)
b )\(-2y\sqrt{2}\)
c )\(5x\sqrt{x}\)
d)\(4y^2\sqrt{3}\)
a) \(\sqrt{27x^2}\)
\(=\sqrt{3^2\cdot3x^2}\)
\(=\left|3x\right|\sqrt{3}\)
\(=3\left|x\right|\sqrt{3}\)
b) \(\sqrt{8xy^2}\)
\(=\sqrt{2^2\cdot2\cdot x\cdot y^2}\)
\(=\left|2y\right|\sqrt{2x}\)
\(=2\left|y\right|\sqrt{2x}\)
c) \(\sqrt{25x^3}\)
\(=\sqrt{5^2\cdot x^2\cdot x}\)
\(=\left|5x\right|\sqrt{x}\)
\(=5\left|x\right|\sqrt{x}\)
d) \(\sqrt{48xy^4}\)
\(=\sqrt{4^2\cdot3x\cdot\left(y^2\right)^2}\)
\(=\left|4y^2\right|\sqrt{3x}\)
\(=4y^2\sqrt{3x}\)
`a, sqrt(27x^2b) = sqrt(3^2. 3.x^2b) = 3|x|sqrt(3b)`.
`b, sqrt(8xy^2) =sqrt(2^2.2xy^2)= 2|y|sqrt(2x)`
`c, sqrt(25x^3d) = sqrt(5^2.x^2.x.d) = 5|x|sqrt(xd)`.
`d, sqrt(48xy^4) = sqrt(4^2.3 . xy^4) = 4y^2sqrt(3x)`.
\(\sqrt{48\cdot45}=12\sqrt{15}\\ \sqrt{225\cdot17}=15\sqrt{17}\\ \sqrt{a^3b^7}=\left|ab^3\right|\sqrt{ab}=ab^3\sqrt{ab}\\ \sqrt{x^5\left(x-3\right)^2}=\left|x^2\left(x-3\right)\right|\sqrt{x}=x^2\left(x-3\right)\sqrt{x}\)
\(\sqrt{48\cdot45}=4\sqrt{3}\cdot3\sqrt{5}=12\sqrt{15}\)
\(\sqrt{225\cdot17}=15\sqrt{17}\)
a: \(a^2\cdot\sqrt{\dfrac{2}{3a}}=a^2\cdot\dfrac{\sqrt{2}}{\sqrt{3}\cdot\sqrt{a}}=\dfrac{a\sqrt{2}}{\sqrt{3}}=\dfrac{a\sqrt{6}}{3}\)
b: \(\dfrac{x-3}{x}\cdot\sqrt{\dfrac{x^3}{9-x^2}}\)
\(=\dfrac{x-3}{x}\cdot\dfrac{x\sqrt{x}}{\sqrt{x-3}\cdot\sqrt{x+3}}\)
\(=\dfrac{\sqrt{x}\cdot\sqrt{x-3}}{\sqrt{x+3}}\)
\(1,ĐKXĐ:x\ge0\\ x\sqrt{3}=-\sqrt{3x^2}\\ \Leftrightarrow3x^2=9x^2\\ \Leftrightarrow6x^2=0\\ \Leftrightarrow x=0\left(tm\right)\)
\(2,ab^2\sqrt{a}=ab^2\sqrt{a}\)
\(3,a\sqrt{\dfrac{b}{a}}=\sqrt{ab}\)
\(\frac{2xy^2}{3ab}\sqrt{\frac{9a^3b^4}{8xy^3}}=\frac{2xy^2}{3ab}\frac{3\sqrt{a^2.a}\sqrt{\left(b^2\right)^2}}{2\sqrt{2xy^2.y}}\)
\(=\frac{2xy^2}{3ab}\frac{3a\sqrt{a}b^2}{2y\sqrt{2xy}}=\frac{6xy^2ab^2\sqrt{a}}{6aby\sqrt{2xy}}=\frac{bxy\sqrt{a}}{\sqrt{2xy}}\)
\(=\frac{bxy\sqrt{2axy}}{2xy}=\frac{b\sqrt{2axy}}{2}\)
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 \(a\ge0;a\ne1\)
\(C=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2.\frac{1}{\left(\sqrt{a}+1\right)^2}=1\)
cảm ơn ạ