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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

13 tháng 7 2021

em cảm ơn ạ yeu

12 tháng 11 2021

5: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)

1) Ta có: \(\left\{{}\begin{matrix}2x-y=3\\x+y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=3\\x=-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-x=-1\end{matrix}\right.\)

Vậy: (x,y)=(1;-1)

2) Ta có: \(A=\dfrac{x+20}{x-4}+\dfrac{2}{\sqrt{x}+2}-\dfrac{6}{\sqrt{x}-2}\)

\(=\dfrac{x+20+2\left(\sqrt{x}-2\right)-6\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+20+2\sqrt{x}-4-6\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)

3 tháng 7 2023

`\sqrt{[27(x-1)^2]/12} +3/2 - (x - 2)\sqrt{[50x^2]/[8(x-2)^2]}`   `(1 < x < 2)`

`=\sqrt{[3(x-1)]^2 .3}/\sqrt{2^2 .3} + 3/2 - (x - 2) \sqrt{(5x)^2 . 2}/\sqrt{[2(x - 2)]^2 . 2}`

`=[3\sqrt{3}|x-1|]/[2\sqrt{3}]+3/2-(x-2)[5\sqrt{2}|x|]/[2\sqrt{2}|x-2|]`

`=[3(x-1)]/2+3/2-[5x(x-2)]/[2(2-x)]`   (Vì `1 < x < 2`)

`=3/2x - 3/2 + 3/2 + 5/2x`

`=4x`

3 tháng 7 2023

em cảm ơn ạ.

 

14 tháng 10 2021

\(a,A=4\sqrt{3}-5\sqrt{3}+2-\sqrt{3}=2-2\sqrt{3}\\ B=\dfrac{x+2\sqrt{x}+8+2\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-4}\\ b,B-\dfrac{1}{2}A=\dfrac{\sqrt{x}}{\sqrt{x}-4}-\dfrac{1}{2}\left(2-2\sqrt{3}\right)=0\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-4}=1+\sqrt{3}\\ \Leftrightarrow\sqrt{x}=\left(1+\sqrt{3}\right)\left(\sqrt{x}-4\right)\Leftrightarrow\sqrt{x}=\sqrt{x}-4\sqrt{3}+\sqrt{3x}-4\\ \Leftrightarrow\sqrt{3x}=4\sqrt{3}+4\\ \Leftrightarrow\sqrt{x}=\dfrac{4\sqrt{3}+4}{\sqrt{3}}\\ \Leftrightarrow\sqrt{x}=\dfrac{12+4\sqrt{3}}{3}\\ \Leftrightarrow x=\dfrac{192+96\sqrt{3}}{9}=\dfrac{64+32\sqrt{3}}{3}\)

14 tháng 10 2021

\(\dfrac{\sqrt{x}}{\sqrt{x}-4}=1-\sqrt{3}\)
Nhỉ???

8 tháng 8 2021

a) \(P=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)

\(=\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\dfrac{2-\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)

\(=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)

\(=\dfrac{4}{4-3}\)

\(=4\)

b) \(Q=\left(1+\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\right).\dfrac{1}{\sqrt{x}}vớix>0,x\ne4\)

\(=\left(\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\sqrt{x}-2}\right).\dfrac{1}{\sqrt{x}}\)

\(=\)\(\dfrac{2\sqrt{x}}{\sqrt{x}-2}.\dfrac{1}{\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{2}{\sqrt{x}-2}\)

1 tháng 2 2022

a, \(A=\dfrac{4\left(3-\sqrt{7}\right)}{2}+2\sqrt{7}=\dfrac{12}{2}=6\)

b, \(B=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\sqrt{x}}\right):\dfrac{2-\sqrt{x}}{x-1}\)

\(=\left(\dfrac{\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{2-\sqrt{x}}{x-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

1 tháng 2 2022

nhờ bạn làm rõ vì sao \(\dfrac{\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{2-\sqrt{x}}{x-1}\) lại bằng \(\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

mình xin cảm ơn

a) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\sqrt{\dfrac{\left(\sqrt{x+1}\right)^2}{\left(\sqrt{x}+1\right)^2}}\)

=\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1};x\ge0\)

b) Ta có: \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)

\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}\)

\(=\dfrac{1}{x-1}\)

1) Ta có: \(\left\{{}\begin{matrix}2x+y=5\\3x-2y=11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x+3y=15\\6x-4y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=-7\\2x+y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\2x=5-y=5-\left(-1\right)=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

2) Ta có: \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{x+3\sqrt{x}+2+2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{x-2\sqrt{x}+2x-4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{1}\)

\(=\dfrac{3x-6\sqrt{x}}{\sqrt{x}-2}\)

\(=3\sqrt{x}\)