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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 \(a\ge0;a\ne1\)
\(C=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2.\frac{1}{\left(\sqrt{a}+1\right)^2}=1\)
\(\sqrt{48\cdot45}=12\sqrt{15}\\ \sqrt{225\cdot17}=15\sqrt{17}\\ \sqrt{a^3b^7}=\left|ab^3\right|\sqrt{ab}=ab^3\sqrt{ab}\\ \sqrt{x^5\left(x-3\right)^2}=\left|x^2\left(x-3\right)\right|\sqrt{x}=x^2\left(x-3\right)\sqrt{x}\)
\(\sqrt{48\cdot45}=4\sqrt{3}\cdot3\sqrt{5}=12\sqrt{15}\)
\(\sqrt{225\cdot17}=15\sqrt{17}\)
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: \(\dfrac{x-3}{x}\cdot\sqrt{\dfrac{x^3}{9-x^2}}\)
\(=\dfrac{x-3}{x}\cdot\dfrac{x\sqrt{x}}{\sqrt{x-3}\cdot\sqrt{x+3}}\)
\(=\dfrac{\sqrt{x}\cdot\sqrt{x-3}}{\sqrt{x+3}}\)
\(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2\cdot\dfrac{2}{x}}=\sqrt{2x}\)
\(x\sqrt{\dfrac{2}{5}}=\sqrt{\dfrac{2}{5}\cdot x^2}=\sqrt{\dfrac{2x^2}{5}}\)
\(\left(x-5\right)\cdot\sqrt{\dfrac{x}{25-x^2}}=\sqrt{\left(x-5\right)^2\cdot\dfrac{x}{-\left(x-5\right)\left(x+5\right)}}=\sqrt{-\dfrac{x\left(x-5\right)}{x+5}}\)
\(x\sqrt{\dfrac{7}{x^2}}=\sqrt{x^2\cdot\dfrac{7}{x^2}}=\sqrt{7}\)
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,\(\sqrt{3\left(x^2-2xy+y^2\right)}=\) \(\sqrt{3\cdot\left(x-y\right)^2}=\left|x-y\right|\sqrt{3}\)
Bài 2:
a: \(=\sqrt{\left(\dfrac{1}{5a}\right)^2}=\dfrac{1}{\left|5a\right|}=\dfrac{-1}{5a}\)
b: \(=\dfrac{1}{3}\cdot15\cdot\left|a\right|=5\left|a\right|\)
\(\frac{1}{x-y}.\sqrt{x^4\left(x^2+y^2-2xy\right)}\)
\(=\frac{1}{x-y}.\sqrt{\left(x^2\right)^2.\left(x-y\right)^2}\)
\(=\frac{1}{x-y}\left(x-y\right)x^2\)
\(=x^2\)
\(\sqrt{18b^3\cdot\left(1-2a\right)^2}\)
\(=3\sqrt{2}\cdot b\sqrt{b}\cdot\left|1-2a\right|\)
\(=3\sqrt{2}\left(2a-1\right)\cdot b\sqrt{b}\)