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a)\(\frac{\sqrt{a-2\sqrt{ab}+b}}{\sqrt{\sqrt{a}-\sqrt{b}}}=\frac{\sqrt{\left(\sqrt{a}-\sqrt{b}\right)^2}}{\sqrt{\sqrt{a}-\sqrt{b}}}=\sqrt{a}-\sqrt{b}\) (vì a > b > 0)
b) \(\frac{\sqrt{x-3}}{\sqrt{\sqrt{x}+\sqrt{3}}}:\frac{\sqrt{\sqrt{x}-\sqrt{3}}}{\sqrt{3}}=\frac{\sqrt{3}.\sqrt{x-3}}{\sqrt{\left(\sqrt{x}+\sqrt{3}\right)\left(\sqrt{x}-\sqrt{3}\right)}}=\frac{\sqrt{3\left(x-3\right)}}{\sqrt{x-3}}=\sqrt{3}\)
c) \(2y^2\sqrt{\frac{x^4}{4y^2}}=2y^2\cdot\frac{x^2}{-2y}=-x^2y\) (vì y < 0)
d) \(\frac{y}{x}\cdot\sqrt{\frac{x^2}{y^4}}=\frac{y}{x}\cdot\frac{x}{y^2}=\frac{1}{y}\)(vì x > 0)
e) \(5xy\cdot\sqrt{\frac{25x^2}{y^6}}=5xy\cdot\frac{-5x}{y^3}=\frac{-25x^2}{y^2}\) (Vì x < 0, y > 0)
\(a,\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}=\left|\sqrt{x}-\sqrt{y}\right|\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)\)
\(=y-x\)
\(b,\frac{3-\sqrt{x}}{x-9}=\frac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\frac{1}{\sqrt{x}+3}\)
\(c,\frac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)
\(d,6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(3-x\right)^2}=6-2x-3+x=3-x\)
\(a,\)\(\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}\)
\(=|\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)|\)
\(=|\sqrt{x}^2-\sqrt{y}^2|\)
\(=|x-y|\)
Vì \(x\le y\)\(\Rightarrow x-y\ge0\)
\(\Rightarrow|x-y|=x-y\)
a) \(\frac{\sqrt{4mn^2}}{\sqrt{20m}}=\sqrt{\frac{4mn^2}{20m}}=\sqrt{\frac{n^2}{5}}=\frac{n}{\sqrt{5}}\)
b) \(\frac{\sqrt{16a^4b^6}}{\sqrt{12a^6b^6}}=\sqrt{\frac{16a^4b^6}{12a^6b^6}}=\sqrt{\frac{4}{3a^2}}=\frac{2}{\sqrt{3}.\left|a\right|}=-\frac{2}{a\sqrt{3}}\)
d) \(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
e) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
a. Ta có:\(\frac{x}{y}\sqrt{\frac{y^2}{x^4}=}\) \(\frac{x}{y}.\frac{\left|y\right|}{x^2}=\frac{x.y}{x^2y}\)\(=\frac{1}{x}\)(Vì \(x\ne0;y>0\))
b \(3x^2\sqrt{\frac{8}{x^2}}=3x^2\frac{2\sqrt{2}}{\left|x\right|}=\frac{6x^2\sqrt{2}}{-x}=-6x\sqrt{2}\)( Vì \(x< 0\))
a/ \(\frac{y}{x}.\left(\sqrt{\frac{x^2}{y^4}}\right)=\frac{y}{x}.\frac{x}{y^2}=\frac{1}{y}\)
b/ \(2y^2.\sqrt{\frac{x^4}{4y^2}}=2y^2.\sqrt{\frac{\left(x^2\right)^2}{\left(-2y\right)^2}}=2y^2.\frac{x^2}{-2y}=-y.x^2\)
c/ \(5xy.\sqrt{\frac{25x^2}{y^6}}=5xy.\sqrt{\frac{\left(-5x\right)^2}{\left(y^3\right)^2}}=5xy.\frac{-5x}{y^3}=\frac{-25x^2}{y^2}\)
d/\(0,2.x^3y^3.\sqrt{\frac{4^2}{\left(x^2y^4\right)^2}}=\frac{1}{5}.x^3y^3.\frac{4}{x^2y^4}=\frac{4x}{5y}\)
Trần Việt Linh sai phần b,c,d r bn
Sửa lại:
b) 2y\(^2\).\(\sqrt{\frac{x^4}{4y^2}}\) với y<0
Ta có : 2y\(^2\).\(\sqrt{\frac{x^4}{4y^2}}\)=2y\(^2\).\(\frac{x^2}{\left|y\right|}\)
Vì y>0 nên |y| = -y.Ta có : 2y\(^2\).\(\frac{x^2}{2\left|y\right|}\)= -2y\(^2\).\(\frac{x^2}{2y}\) = -2x\(^2\)y
c) 5xy.\(\sqrt{\frac{25x^2}{y^6}}\) với x<0,y>0
Ta có :5xy\(\sqrt{\frac{25x^2}{y^6}}\)=5xy.\(\frac{5\left|x\right|}{y^3}\) ( y>0)
Vì x<0 nên |x| =-x .Ta có : 5xy.\(\frac{5\left|x\right|}{y^3}\)= -5xy.\(\frac{5x}{y^3}\) =\(\frac{-25x^2}{y^2}\)
d) 0,,2x\(^3\)y\(^3\).\(\sqrt{\frac{16}{x^4y^8}}\) với x#o,y#0
Ta có: 0,2x\(^3\)y\(^3\)\(\frac{4}{x^2y^4}\)=\(\frac{0,8x}{y}\) ( vì #0,y#0)