Question

Cho x,y>0 thỏa mãn hệ thức : \(\sqrt{x}(\sqrt{x}+\sqrt{y})=3\sqrt{y}\left( \sqrt{x}+5\sqrt{y} \right) \)     (1)

Hãy tính giá trị biểu thức \(E=\frac{2x+\sqrt{xy}+3y}{x+\sqrt{xy}-y}\)

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thưởng 3 tick !!!

Answer 1

Ta có :\(\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)=3\sqrt{y}\left(\sqrt{x}+5\sqrt{y}\right)\)

\(\Leftrightarrow x+\sqrt{xy}-3\sqrt{xy}-15y=0\)

\(\Leftrightarrow x-2\sqrt{xy}+y-16y=0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2-\left(4\sqrt{y}\right)^2=0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}-4\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}+4\sqrt{y}\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-5\sqrt{y}\right)\left(\sqrt{x}+3\sqrt{y}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-5\sqrt{y}=0\\\sqrt{x}+3\sqrt{y}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=5\sqrt{y}\\\sqrt{x}=-3\sqrt{y}\end{cases}}\)

\(\Leftrightarrow\sqrt{x}=5\sqrt{y}\)(do x,y>0)

\(\Leftrightarrow x=25y\)(*)

Thay (*) vào biểu thức E ta được: \(E=\frac{2.25y+\sqrt{25y.y}+3y}{25y+\sqrt{25y.y}-y}=\frac{50y+5y+3y}{25y+5y-y}=\frac{58y}{29y}=2\)

Vậy giá trị của biểu thức E là 2.

Answer 2

ta có:\(\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)=3\sqrt{y}\left(\sqrt{x}+5\sqrt{y}\right)\Leftrightarrow x-2\sqrt{xy}-3y-15y=0\Leftrightarrow\)

\(\left(\sqrt{x}-\sqrt{y}\right)^2-\left(4\sqrt{y}\right)^2=0\Leftrightarrow\left(\sqrt{x}+3\sqrt{y}\right)\left(\sqrt{x}-5\sqrt{y}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+3\sqrt{y}=0\\\sqrt{x}-5\sqrt{y}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=-3\sqrt{y}\left(loai\left(vi-x,y>0\right)\right)\\\sqrt{x}=5\sqrt{y}\end{cases}}}\)

thay \(\sqrt{x}=5\sqrt{y}\) vào E ta có:

\(E=\frac{2\left(5\sqrt{y}\right)^2+5\sqrt{y.y}+3y}{\left(\sqrt{5y}\right)^2+5\sqrt{y.y}-y}=\frac{y\left(50+5+3\right)}{y\left(25+5-1\right)}=2\)

vậy E =2