Rút gọn biểu thức A
A=(\(\dfrac{4\sqrt{y}}{2+\sqrt{y}}+\dfrac{8y}{4-y}\)) : (\(\dfrac{\sqrt{y}-1}{y-2\sqrt{y}}-\dfrac{2}{\sqrt{y}}\))
với y>0; y≠4; y≠9
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12 tháng 11 2021
5: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)
31 tháng 12 2022
a: \(A=\dfrac{4y-8\sqrt{y}-8y}{y-4}:\dfrac{\sqrt{y}-1-2\sqrt{y}+4}{\sqrt{y}\left(\sqrt{y}-2\right)}\)
\(=\dfrac{-4\sqrt{y}\left(\sqrt{y}+2\right)}{y-4}\cdot\dfrac{\sqrt{y}\left(\sqrt{y}-2\right)}{-\sqrt{y}+3}\)
\(=\dfrac{4y}{\sqrt{y}-3}\)
b: Để A=-2 thì \(4y=-2\sqrt{y}+6\)
=>\(4y+2\sqrt{y}-6=0\)
=>y=1
TN
10 tháng 8 2018
\(A=\left(\dfrac{4\sqrt{y}}{2+\sqrt{y}}+\dfrac{8y}{4-y}\right):\left(\dfrac{\sqrt{y}-1}{y-2\sqrt{y}}-\dfrac{2}{\sqrt{y}}\right)\\ =\left(\dfrac{4\sqrt{y}.\left(2-\sqrt{y}\right)+8y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}\right):\left(\dfrac{\sqrt{y}-1-2\left(\sqrt{y}-2\right)}{\sqrt{y}\left(\sqrt{y}-2\right)}\right)\\ =\left(\dfrac{4\sqrt{y}\left(2+\sqrt{y}\right)}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}\right):\left(\dfrac{3-\sqrt{y}}{\sqrt{y}\left(\sqrt{y}-2\right)}\right)\\ =.\dfrac{4\sqrt{y}.\left(-\sqrt{y}\right)\left(2-\sqrt{y}\right)}{\left(2-\sqrt{y}\right)\left(3-\sqrt{y}\right)}\\ =\dfrac{-4y}{3-\sqrt{y}}\)
Ta có:
\(A=\dfrac{-4y}{3-\sqrt{y}}=-2\Rightarrow-4y=-6+2\sqrt{y}\Rightarrow-4y+4\sqrt{y}-6\sqrt{y}+6=0\\ \Rightarrow-4\sqrt{y}\left(\sqrt{y}-1\right)-6\left(\sqrt{y}-1\right)=0\\ \Rightarrow\left(\sqrt{y}-1\right)\left(-4\sqrt{y}-6\right)=0\Rightarrow\sqrt{y}-1=0\Rightarrow y=1\)
2 tháng 10 2021
a: ĐKXĐ: \(\left\{{}\begin{matrix}y\ge0\\y\ne1\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{1}{1-\sqrt{y}}+\dfrac{1}{1+\sqrt{y}}\right):\left(\dfrac{1}{1-\sqrt{y}}-\dfrac{1}{1+\sqrt{y}}\right)+\dfrac{1}{1-\sqrt{y}}\)
\(=\dfrac{1+\sqrt{y}+1-\sqrt{y}}{\left(1-\sqrt{y}\right)\left(1+\sqrt{y}\right)}:\dfrac{1+\sqrt{y}-1+\sqrt{y}}{\left(1-\sqrt{y}\right)\left(1+\sqrt{y}\right)}+\dfrac{1}{1-\sqrt{y}}\)
\(=\dfrac{2}{2\sqrt{y}}-\dfrac{1}{\sqrt{y}-1}\)
\(=\dfrac{\sqrt{y}-1-\sqrt{y}}{\sqrt{y}\left(\sqrt{y}-1\right)}\)
\(=\dfrac{-1}{\sqrt{y}\left(\sqrt{y}-1\right)}\)
8 tháng 7 2021
a) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\sqrt{\dfrac{\left(\sqrt{x+1}\right)^2}{\left(\sqrt{x}+1\right)^2}}\)
=\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1};x\ge0\)
8 tháng 7 2021
b) Ta có: \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}\)
\(=\dfrac{1}{x-1}\)
\(A=\left(\dfrac{4\sqrt{y}}{2+\sqrt{y}}+\dfrac{8y}{4-y}\right):\left(\dfrac{\sqrt{y}-1}{y-2\sqrt{y}}-\dfrac{2}{\sqrt{y}}\right)\)
\(A=\dfrac{4\sqrt{y}\left(2-\sqrt{y}\right)+8y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}:\dfrac{\sqrt{y}-1-2\left(\sqrt{y}-2\right)}{\sqrt{y}\left(\sqrt{y}-2\right)}\)
\(A=\dfrac{8\sqrt{y}+4y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}.\dfrac{\sqrt{y}\left(\sqrt{y}-2\right)}{-\sqrt{y}+3}\)
\(A=\dfrac{4\sqrt{y}}{2-\sqrt{y}}.\dfrac{\sqrt{y}\left(2-\sqrt{y}\right)}{\sqrt{y}-3}\)
\(A=\dfrac{4y}{\sqrt{y}-3}\)
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