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19 tháng 8 2021
\(2\sqrt{3x}-\sqrt{75x}\)
\(=2\sqrt{3x}-5\sqrt{3x}\)
\(=-3\sqrt{3x}\)
7 tháng 9 2021
\(A=\sqrt[3]{\left(4-2\sqrt[3]{3}\right)\left(\sqrt[3]{3}-1\right)}\)
\(=\sqrt[3]{4\sqrt[3]{3}-4-2\sqrt[3]{9}+2\sqrt[3]{3}}\)
\(=\sqrt[3]{4-2\sqrt[3]{9}+6\sqrt[3]{3}}\)
23 tháng 10 2021
\(\left(3+\dfrac{\sqrt{6}-\sqrt{10}}{\sqrt{5}-\sqrt{3}}\right)\left(3+\dfrac{2\sqrt{5}+\sqrt{6}}{\sqrt{10}+\sqrt{3}}\right)\)
\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)
=9-2
=7
4 tháng 11 2021
a: \(=\sqrt{26}-5-\sqrt{26}=-5\)
b: \(=\sqrt{7}-\sqrt{2}-\sqrt{7}=-\sqrt{2}\)
4 tháng 7 2021
\(C=\sqrt{\dfrac{x-6\sqrt{x}+9}{x+6\sqrt{x}+9}}=\sqrt{\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+3\right)^2}}=\dfrac{\left|\sqrt{x}-3\right|}{\sqrt{x}+3}\)
Vì \(x\ge9\Rightarrow\sqrt{x}\ge3\Leftrightarrow\sqrt{x}-3\ge0\)
\(\Rightarrow C=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(D=\dfrac{x-1}{\sqrt{y}-1}.\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\) (\(x;y\ne1;y\ge0\))
\(=\dfrac{x-1}{\sqrt{y}-1}.\dfrac{\sqrt{\left(\sqrt{y}-1\right)^2}}{\left(x-1\right)^2}=\dfrac{\left|\sqrt{y}-1\right|}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
TH1: \(\sqrt{y}-1>0\Leftrightarrow y>1\)
\(\Rightarrow D=\dfrac{\sqrt{y}-1}{\left(\sqrt{y}-1\right)\left(x-1\right)}=\dfrac{1}{x-1}\)
TH2:\(\sqrt{y}-1< 0\Leftrightarrow0\le y< 1\)
\(\Rightarrow D=\dfrac{-\left(\sqrt{y}-1\right)}{\left(\sqrt{y}-1\right)\left(x-1\right)}=\dfrac{-1}{x-1}\)
Vậy...
\(E=\dfrac{1}{2x-1}.\sqrt{5x^4\left(1-4x+4x^2\right)}\)
\(=\dfrac{1}{2x-1}\sqrt{5x^4\left(2x-1\right)^2}=\dfrac{\sqrt{5}x^2\left|2x-1\right|}{2x-1}\)
TH1: \(2x-1>0\Leftrightarrow x>\dfrac{1}{2}\)
\(\Rightarrow E=\dfrac{\sqrt{5}x^2\left(2x-1\right)}{2x-1}=\sqrt{5}x^2\)
TH2:\(2x-1< 0\Leftrightarrow x< \dfrac{1}{2}\)
\(\Rightarrow E=\dfrac{-\sqrt{5}x^2\left(2x-1\right)}{2x-1}=-\sqrt{5}x^2\)
Vậy...
H
4 tháng 7 2021
c)
\(C=\sqrt{\dfrac{x-6\sqrt{x}+9}{x+6\sqrt{x}+9}}=\sqrt{\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+3\right)^2}}=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
d)
\(D=\dfrac{x-1}{\sqrt{y}-1}.\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}=\dfrac{x-1}{\sqrt{y}-1}.\dfrac{\left|\sqrt{y}-1\right|}{\left(x-1\right)^2}=\dfrac{\left|\sqrt{y}-1\right|}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
e)
\(E=\dfrac{1}{2x-1}.\sqrt{5x^4\left(1-4x+4x^2\right)}=\dfrac{1}{2x-1}.\sqrt{5x^4.\left(2x-1\right)^2}=\dfrac{1}{2x-1}.\sqrt{5}x^2.\left|2x-1\right|\)
TN
2
19 tháng 8 2021
\(\sqrt{\dfrac{x}{y}}+\sqrt{xy}+\dfrac{x}{y}\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{x}{y}}\cdot\sqrt{y^2}+\sqrt{\dfrac{x}{y}}\cdot\sqrt{\dfrac{x}{y}\cdot\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}\cdot\left(1+y+1\right)=\sqrt{\dfrac{x}{y}}\cdot\left(y+2\right)\)
19 tháng 8 2021
Ta có: \(\sqrt{\dfrac{x}{y}}+\sqrt{xy}+\dfrac{x}{y}\cdot\sqrt{\dfrac{y}{x}}\)
\(=\dfrac{\sqrt{x}}{\sqrt{y}}+\dfrac{\sqrt{x}}{\sqrt{y}}+\sqrt{xy}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{y}}+\dfrac{y\sqrt{x}}{\sqrt{y}}\)
\(=\dfrac{2\sqrt{x}+y\sqrt{x}}{\sqrt{y}}\)
c: \(C=\dfrac{\left(\sqrt{a}-2\right)\left(a+2\sqrt{a}+4\right)+2\sqrt{a}\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{a+2\sqrt{a}+4+2\sqrt{a}}{\sqrt{a}+2}=\sqrt{a}+2\)
d: \(=\dfrac{1}{2a-1}\cdot a^2\sqrt{5}\cdot\left|2a-1\right|=\pm a^2\sqrt{5}\)
e: \(=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\sqrt{3}\left|x+y\right|}{2}\)
\(=\pm\dfrac{\sqrt{3}}{x-y}\)