thực hiện phép tính
\(\sqrt{50^3y^5}-\frac{2y^2}{x^2}\sqrt{32x^7y}+\frac{3xy}{2}\sqrt{2xy^3},x>0,y>0\)
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20 tháng 3 2019
b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
28 tháng 7 2022
a: \(=\dfrac{\sqrt{m}\left(m+4n-4\sqrt{mn}\right)}{\sqrt{mn}\left(\sqrt{m}-2\sqrt{n}\right)}\)
\(=\dfrac{1}{\sqrt{n}}\cdot\left(\sqrt{m}-2\sqrt{n}\right)\)
b: \(=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
c: \(=\sqrt{5^2\cdot2\cdot x^2y^4\cdot xy}-\dfrac{2y^2}{x^2}\cdot4\sqrt{2}\cdot x^3\sqrt{xy}+\dfrac{3}{2}xy\cdot\sqrt{2}\cdot y\cdot\sqrt{xy}\)
\(=5xy^2\sqrt{2xy}-8\sqrt{2xy}xy^2+\dfrac{3}{2}xy^2\cdot\sqrt{2xy}\)
\(=-\dfrac{3}{2}\sqrt{2xy}\)
d: \(=\left(x+2\right)\cdot\dfrac{\sqrt{2x-3}}{\sqrt{x+2}}=\sqrt{\left(2x-3\right)\left(x+2\right)}\)
T
17 tháng 2 2018
Xét VT của (1):
\(3VT\)
\(=\sqrt{5x^2+2xy+2y^2}.\sqrt{2^2+2^2+1^2}+\sqrt{2x^2+2xy+5y^2}.\sqrt{2^2+2^2+1^2}\)
\(=\sqrt{\left(x+y\right)^2+4x^2+y^2}.\sqrt{2^2+2^2+1^2}+\sqrt{\left(x+y\right)^2+x^2+4y^2}.\sqrt{2^2+2^2+1^2}\)
\(\ge\left[2\left(x+y\right)+4x+y\right]+\left[2\left(x+y\right)+x+4y\right]=9x+9y\)
\(\Rightarrow VT\ge3x+3y=VT\)
Đẳng thức xảy ra \(\Leftrightarrow...\Leftrightarrow x=y\)
Sau đó thay \(y=x\) vào pt (2) ta được:
\(\sqrt{3x+1}+2\sqrt[3]{19x+8}=2x^2+x+5\)
\(\Leftrightarrow\left(2x^2-\sqrt{3x+1}\right)+\left(x-5-2\sqrt[3]{19x+8}\right)=0\)
\(\Leftrightarrow\dfrac{4x^2-3x-1}{2x^2+\sqrt{3x+1}}+\dfrac{\left(x+5\right)^3-8\left(19x+8\right)}{\left(x-5\right)^2+2\left(x-5\right)\sqrt[3]{19x+8}+4\sqrt[3]{\left(19x+8\right)^2}}=0\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(4x+1\right)}{2x^2+\sqrt{3x+1}}+\dfrac{ \left(x-1\right)\left(x^2+16x-61\right)}{\left(x-5\right)^2+2\left(x-5\right)\sqrt[3]{19x+8}+4\sqrt[3]{\left(19x+8\right)^2}}=0\)
\(\Leftrightarrow\left(x-1\right)\left[\dfrac{4x+1}{2x^2+\sqrt{3x+1}}+\dfrac{x^2+16x-61}{\left(x-5\right)^2+2\left(x-5\right)\sqrt[3]{19x+8}+4\sqrt[3]{\left(19x+8\right)^2}}\right]=0\)
\(\Leftrightarrow x=1\Rightarrow y=1\)
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Kết quả rất lẻ : \(\frac{-16y^2\sqrt{x^7y}+3x^3y\sqrt{xy^3}+500x^2\sqrt{y^5}}{\sqrt{2}x^2}\)