rút gọn
a, \(\dfrac{x}{y}\sqrt{\dfrac{x^2}{y^4}}\) với x>0, y khác 0
b, \(2y^2\sqrt{\dfrac{x^4}{4y^2}}\) với y<0
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1 tháng 10 2023
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) \(ab^2\sqrt{\dfrac{3}{a^2b^4}}\)
\(=ab^2\dfrac{\sqrt{3}}{\sqrt{\left(ab^2\right)^2}}\)
\(=ab^2\cdot\dfrac{\sqrt{3}}{ab^2}\)
\(=\sqrt{3}\)
T
1 tháng 10 2023
\(a,\dfrac{y}{x}\cdot\sqrt{\dfrac{x^2}{y^4}}\left(y\ge0;x,y\ne0\right)\) (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)
\(=ab^2\cdot\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}\)
\(=\dfrac{ab^2\sqrt{3}}{\sqrt{\left(ab^2\right)^2}}\)
\(=\dfrac{ab^2\sqrt{3}}{ab^2}\)
\(=\sqrt{3}\)
#\(Toru\)
1 tháng 10 2023
a) \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}\)
\(=\sqrt{\dfrac{2a\cdot3a}{3\cdot8}}\)
\(=\sqrt{\dfrac{6a^2}{24}}\)
\(=\sqrt{\dfrac{a^2}{4}}\)
\(=\dfrac{\sqrt{a^2}}{\sqrt{4}}\)
\(=\dfrac{a}{2}\)
b) \(\sqrt{3a}\cdot\sqrt{\dfrac{52}{a}}\)
\(=\sqrt{3a\cdot\dfrac{52}{a}}\)
\(=\sqrt{3\cdot52}\)
\(=\sqrt{13\cdot3\cdot4}\)
\(=2\sqrt{39}\)
c) \(2y^2\cdot\sqrt{\dfrac{x^4}{4y^2}}\)
\(=2y^2\cdot\dfrac{\sqrt{\left(x^2\right)^2}}{\sqrt{\left(2y\right)^2}}\)
\(=2y^2\cdot\dfrac{x^2}{-2y}\)
\(=\dfrac{2y^2\cdot x^2}{-2y}\)
\(=-x^2y\)
6 tháng 2 2022
a: \(=\dfrac{\left|x+2\right|}{x-1}\)
b: \(=x-2y-\left|x-2y\right|\)\(=\left[{}\begin{matrix}x-2y-x+2y=0\\x-2y+x-2y=2x-4y\end{matrix}\right.\)
c: \(=\dfrac{\left|x+2\right|}{\left(x+2\right)\left(x-2\right)}=\pm\dfrac{1}{x-2}\)
12 tháng 7 2021
a) Ta có: \(\dfrac{x^2}{y^2}:\sqrt{\dfrac{x^2}{y^4}}\)
\(=\dfrac{x^2}{y^2}:\dfrac{x}{y^2}\)
=x
b) Ta có: \(\sqrt{\dfrac{27\left(x-1\right)^2}{12}}+\dfrac{3}{2}-\left(x-2\right)\sqrt{\dfrac{50x^2}{8\left(x-2\right)^2}}\)
\(=\sqrt{\dfrac{9}{4}}\cdot\sqrt{\left(x-1\right)^2}+\dfrac{3}{2}-\left(x-2\right)\cdot\sqrt{\dfrac{25}{4}}\cdot\sqrt{\dfrac{x^2}{\left(x-2\right)^2}}\)
\(=\dfrac{3}{2}\cdot\left(x-1\right)+\dfrac{3}{2}-\left(x-2\right)\cdot\dfrac{5}{2}\cdot\dfrac{x}{2-x}\)
\(=\dfrac{3}{2}x-\dfrac{3}{2}+\dfrac{3}{2}-\dfrac{5}{2}\left(x-2\right)\cdot\dfrac{-x}{x-2}\)
\(=\dfrac{3}{2}x+\dfrac{5}{2}\cdot\left(x\right)\)
=4x
N
19 tháng 11 2021
\(1,A=\dfrac{2x+1-x}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\left(x-\sqrt{x}-2\right)\\ A=\dfrac{\left(x+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\left(x+1\right)\left(\sqrt{x}-2\right)}{x-\sqrt{x}+1}\\ 2,\Leftrightarrow\left\{{}\begin{matrix}2a-b=1\\a-b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=-3\end{matrix}\right.\Leftrightarrow y=-x-3\)
T
19 tháng 4 2021
(Vì x > 0 nên |x| = x; y2 > 0 với mọi y ≠ 0)
(Vì x2 ≥ 0 với mọi x; và vì y < 0 nên |2y| = – 2y)
(Vì x < 0 nên |5x| = – 5x; y > 0 nên |y3| = y3)
(Vì x2y4 = (xy2)2 > 0 với mọi x ≠ 0, y ≠ 0)
12 tháng 11 2021
5: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)
R
1
17 tháng 1 2022
\(=\dfrac{2}{x+y}\cdot\dfrac{\sqrt{3}\left(x+y\right)}{2}=\sqrt{3}\)
\(a,=\dfrac{x}{y}\cdot\dfrac{\left|x\right|}{y^2}=\dfrac{x^2}{y^3}\\ b,=2y^2\cdot\dfrac{x^2}{\left|2y\right|}=\dfrac{2x^2y^2}{-2y}=-x^2y\)