Rút gọn biểu thức sau:
\(\frac{x}{x-y}-\frac{1}{x-y}-\frac{1-y}{y-x}\)
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BT
1 tháng 12 2017
a/ \(\frac{7x-14y}{x^2-4y^2}=\frac{7\left(x-2y\right)}{x^2-\left(2y\right)^2}=\frac{7\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{7}{x+2y}.\)
b/ \(\frac{1-\frac{2y}{x}+\frac{y^2}{x^2}}{\frac{1}{x}-\frac{1}{y}}=\frac{\frac{x^2-2xy+y^2}{x^2}}{\frac{y-x}{xy}}=\frac{\left(x-y\right)^2}{x^2}.\frac{xy}{-\left(x-y\right)}=-\frac{y\left(x-y\right)}{x}=\frac{y\left(y-x\right)}{x}\)
15 tháng 8 2023
a: \(A=\dfrac{x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{2}}+y^{\dfrac{1}{3}}\cdot x^{\dfrac{1}{2}}}{x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}}=\dfrac{x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{3}}\left(x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}\right)}{x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}}=x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{3}}=\left(xy\right)^{\dfrac{1}{3}}\)
b: \(B=\dfrac{x^{3+\sqrt{3}}}{y^2}\cdot\dfrac{x^{-\sqrt{3}-1}}{y^{-2}}=\dfrac{x^{3+\sqrt{3}-\sqrt{3}-1}}{y^{2-2}}=x^2\)
HQ
Hà Quang Minh
Giáo viên
23 tháng 8 2023
\(A=\dfrac{x^{\dfrac{5}{4}}y+xy^{\dfrac{5}{4}}}{\sqrt[4]{x}+\sqrt[4]{y}}\\ =\dfrac{xy\left(x^{\dfrac{1}{4}}+y^{\dfrac{1}{4}}\right)}{x^{\dfrac{1}{4}}+y^{\dfrac{1}{4}}}\\ =xy\)
\(B=\left(\sqrt[7]{\dfrac{x}{y}\sqrt[5]{\dfrac{y}{x}}}\right)^{\dfrac{35}{4}}\\= \left(\sqrt[7]{\dfrac{x}{y}\cdot\left(\dfrac{x}{y}\right)^{-\dfrac{1}{5}}}\right)^{\dfrac{35}{4}}\\ =\left(\sqrt[7]{\left(\dfrac{x}{y}\right)^{\dfrac{4}{5}}}\right)^{\dfrac{35}{4}}\\ =\left[\left(\dfrac{x}{y}\right)^{\dfrac{4}{35}}\right]^{\dfrac{35}{4}}\\ =\left(\dfrac{x}{y}\right)^{\dfrac{4}{35}\cdot\dfrac{35}{4}}\\ =\left(\dfrac{x}{y}\right)^1\\ =\dfrac{x}{y}\)
6 tháng 7 2019
\(A=\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x^2-2x+1\right)^2}}\)
\(=\frac{x-1}{\sqrt{y}-1}.\frac{\sqrt{\left(y-2\sqrt{y}+1\right)^2}}{\sqrt{\left(x^2-2x+1\right)^2}}\)
\(=\frac{x-1}{\sqrt{y}-1}.\frac{|y-2\sqrt{y}+1|}{|(x^2-2x+1)|}\)
\(=\frac{x-1}{\sqrt{y}-1}.\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
6 tháng 7 2019
\(đkxđ\Leftrightarrow\hept{\begin{cases}y\ge0\\x-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}y\ge0\\x\ne1\end{cases}}}\)
\(A=\frac{x-1}{\sqrt{y}-1}\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x^2-2x+1\right)^2}}\)
\(=\frac{x-1}{\sqrt{y}-1}\sqrt{\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}}\)
\(=\frac{\left(x-1\right)|\sqrt{y}-1|}{\left(\sqrt{y}-1\right)\left(x-1\right)}=\frac{|\sqrt{y}-1|}{\left(\sqrt{y}-1\right)}\)
TH1 : \(y>1\Rightarrow\sqrt{y}>1\Rightarrow\sqrt{y}-1>0\)
\(\Rightarrow|\sqrt{y}-1|=\sqrt{y}-1\)
\(\Rightarrow A=\frac{\sqrt{y}-1}{\sqrt{y}-1}=1\)
Th2 : \(0< y< 1\Rightarrow\sqrt{y}< 1\Rightarrow\sqrt{y}-1< 0\)
\(\Rightarrow|\sqrt{y}-1|=-\left(\sqrt{y}-1\right)\)
\(\Rightarrow A=\frac{-\left(\sqrt{y}-1\right)}{\sqrt{y}-1}=-1\)
KL : Nếu \(0< y< 1\Rightarrow A=-1\)
Nếu \(y>1\Rightarrow A=1\)
15 tháng 8 2023
\(=\dfrac{xy\left(x^{\dfrac{1}{2}}+y^{\dfrac{1}{2}}\right)}{x^{\dfrac{1}{2}}+y^{\dfrac{1}{2}}}=xy\)
23 tháng 8 2023
\(A=\dfrac{x^{\dfrac{3}{2}}y+xy^{\dfrac{3}{2}}}{\sqrt{x}+\sqrt{y}}=\left(x+y\right).\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\).
\(\frac{x}{x-y}-\frac{1}{x-y}-\frac{1-y}{y-x}=\frac{x}{x-y}-\frac{1}{x-y}+\frac{y-1}{x-y}\)
\(=\frac{x-1+y-1}{x-y}=\frac{x+y-2}{x-y}\)
\(\frac{x}{x-y}-\frac{1}{x-y}-\frac{1-y}{y-x}\)
ĐKXĐ : \(\hept{\begin{cases}x,y\ne0\\x\ne y\end{cases}}\)
\(=\frac{x}{x-y}-\frac{1}{x-y}-\frac{y-1}{x-y}\)
\(=\frac{x-1-y+1}{x-y}\)
\(=\frac{x-y}{x-y}=1\)