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a)\(=-\sqrt{\left(\frac{a}{b}\right)^2\cdot\frac{b}{a}}\)
\(=-\sqrt{\frac{a^2}{b^2}\cdot\frac{b}{a}}\)
\(=-\sqrt{\frac{a}{b}}\)
\(\left(x-5\right)\sqrt{\frac{3}{25-x^2}}=\sqrt{\left(x-5\right)^2}\sqrt{\frac{3}{\left(5-x\right)\left(x+5\right)}}=\sqrt{\left(5-x\right)^2.\frac{3}{\left(5-x\right)\left(x+5\right)}}=\sqrt{\frac{3\left(5-x\right)}{x+5}}\)
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) \(x\sqrt{\frac{1}{x}}=\sqrt{x^2\cdot\frac{1}{x}}=\sqrt{\frac{x^2}{x}}=\sqrt{x}\)( với x > 0 )
b) \(x\sqrt{\frac{-1}{x}}=-\sqrt{x^2\cdot\frac{1}{x}}=-\sqrt{\frac{x^2}{x}}=-\sqrt{x}\)( với x < 0 )
a, \(\sqrt{\frac{\left(x-y\right)^2}{x^2}\cdot\frac{x}{x-y}}=\) \(\frac{x-y}{x}\)
b. \(\sqrt{\frac{\left(x+y\right)^2}{\left(x-y\right)^2}\cdot\frac{x-y}{x+y}}=\sqrt{\frac{x+y}{x-y}}\)
c.\(\sqrt{\frac{x^4}{\left(x-5\right)^2}\cdot\frac{x-5}{3x}}=\sqrt{\frac{x^3}{3\left(x-5\right)}}\)
\(-2\sqrt{-a}=\sqrt{\left(-2\right)^2\cdot-a}=\sqrt{-4a}\)
dài v nhg thui cố làm v
a)\(\sqrt{4x^2}-20x+25+2x=5\)
=> \(2x-18x+20=0\)
=> \(-16x+20=0\)
=> \(-4x+5=0\)
=> \(-4x=-5\)
=> \(x=\dfrac{5}{4}\)
vậy........................................................
d) \(\sqrt{x-2}\cdot\sqrt{x-1}=\sqrt{x-1-1}\)
cau này đề sai
ok baby
\(\frac{1}{2x-1}\sqrt{5-20x+20x^2}=\frac{1}{2x-1}\sqrt{5.\left(1-4x+4x^2\right)}\)
\(=\frac{1}{2x-1}\sqrt{5.\left(1-2x\right)^2}=\sqrt{\frac{1}{\left(2x-1\right)^2}}\sqrt{5.\left(2x-1\right)^2}\)(x>1/2)
\(=\sqrt{\frac{1}{\left(2x-1\right)^2}.5.\left(2x-1\right)^2}=\sqrt{5}\)
thanks nhìu