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a, Ta có : \(\frac{y}{x}.\sqrt{\frac{x^2}{y^4}}=\frac{y}{x}.\frac{x}{y^2}=\frac{1}{y}\)
b , Ta có : \(5xy\sqrt{\frac{x^2}{y^6}}=5xy\frac{x}{y^3}=\frac{5x^2}{y^2}\)
c, Ta có : \(0,2x^3y^3\sqrt{\frac{16}{x^4y^8}}=0,2x^3y^3.\frac{4}{x^2y^4}=\frac{0,8x}{y}\)
\(5xy.\sqrt{\frac{25x^2}{y^6}}=5xy.\sqrt{\frac{5^2x^2}{\left(y^3\right)^2}}=5xy.\sqrt{\frac{\left(5x\right)^2}{\left(y^3\right)^2}}5xy.\sqrt{\left(\frac{5x}{y^3}\right)^2}=5xy.\frac{5x}{y^3}=\frac{5^2x^2}{y^2}=\frac{\left(5x\right)^2}{y^2}=\left(\frac{5x}{y}\right)^2\)
Chúc bạn học tốt
5xy.\(\sqrt{\frac{25x^2}{y^6}}\)
=5xy.\(\frac{\left|5x\right|}{\left|y^3\right|}\){x<0 nên |5x|=-5x
=\(\orbr{\begin{cases}5xy.\frac{-5x}{y^3}\\5xy.\frac{-5x}{-y^3}\end{cases}}\)
=\(\orbr{\begin{cases}\frac{-25x^2}{y^3}\\\frac{25x^2}{y^3}\end{cases}}\)
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)
\(Q=\frac{x-y}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{x^3}-\sqrt{y^3}}{x-y}\)
\(Q=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-y\right)-x\sqrt{x}+y\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(Q=\frac{x\sqrt{x}-y\sqrt{x}+x\sqrt{y}-y\sqrt{y}-x\sqrt{x}+y\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(Q=\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(Q=\frac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(R=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(R=\left[\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right].\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(R=\left(1+\sqrt{a}+a\right).\frac{\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)^2.\left(1+\sqrt{a}\right)^2}\)
\(=\left(1+\sqrt{a}\right)^2.\frac{1}{\left(1+\sqrt{a}\right)^2}=1\)
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)
aizzzz bài này giải rồi mà taaa
Lướt xuống là thấy
Học tốt!!!!!
a) \(\dfrac{y}{x}\cdot\sqrt{\dfrac{x^2}{y^4}}\)
\(=\dfrac{y}{x}\cdot\dfrac{\sqrt{x^2}}{\sqrt{\left(y^2\right)^2}}\)
\(=\dfrac{y}{x}\cdot\dfrac{x}{y^2}\)
\(=\dfrac{1}{y}\)
b) \(\dfrac{5}{2}x^3y^3\cdot\sqrt{\dfrac{16}{x^4y^8}}\)
\(=\dfrac{5}{2}x^3y^3\cdot\dfrac{\sqrt{16}}{\sqrt{\left(x^2y^4\right)^2}}\)
\(=\dfrac{5}{2}x^3y^3\cdot\dfrac{4}{x^2y^4}\)
\(=\dfrac{20x^3y^3}{2x^2y^4}\)
\(=\dfrac{10x}{y}\)
c) \(ab^2\sqrt{\dfrac{3}{a^2b^4}}\)
\(=ab^2\dfrac{\sqrt{3}}{\sqrt{\left(ab^2\right)^2}}\)
\(=ab^2\cdot\dfrac{\sqrt{3}}{ab^2}\)
\(=\sqrt{3}\)
\(a,\dfrac{y}{x}\cdot\sqrt{\dfrac{x^2}{y^4}}\left(y\ge0;x,y\ne0\right)\) (sửa đề)
\(=\dfrac{y}{x}\cdot\dfrac{\sqrt{x^2}}{\sqrt{y^4}}\)
\(=\dfrac{y}{x}\cdot\dfrac{x}{\sqrt{\left(y^2\right)^2}}\)
\(=\dfrac{y}{x}\cdot\dfrac{x}{y^2}\)
\(=\dfrac{1}{y}\)
\(---\)
\(b,\dfrac{5}{2}x^3y^3\cdot\sqrt{\dfrac{16}{x^4y^8}}\left(x,y\ne0\right)\)
\(=\dfrac{5}{2}x^3y^3\cdot\dfrac{\sqrt{16}}{\sqrt{x^4y^8}}\)
\(=\dfrac{5x^3y^3}{2}\cdot\dfrac{4}{x^2y^4}\)
\(=\dfrac{5x\cdot2}{y}\)
\(=\dfrac{10x}{y}\)
\(---\)
\(c,ab^2\sqrt{\dfrac{3}{a^2b^4}}\left(a>0;b\ne0\right)\) (sửa đề)
\(=ab^2\cdot\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}\)
\(=\dfrac{ab^2\sqrt{3}}{\sqrt{\left(ab^2\right)^2}}\)
\(=\dfrac{ab^2\sqrt{3}}{ab^2}\)
\(=\sqrt{3}\)
#\(Toru\)
\(5xy\sqrt{\frac{x^2}{y^6}}=5\sqrt{\frac{x^4y^2}{y^6}}=5\sqrt{\frac{x^4}{y^4}}=5\left|\frac{x^2}{y^2}\right|=-5\)
\(5xy\sqrt{\frac{x^2}{y^6}}=5\sqrt{\frac{x^4y^2}{y^6}}=5\sqrt{\frac{x^4}{y^4}}=5\)