2k6 thì dạng này EZ quá còn gì:)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-5\sqrt{y}=0\\\sqrt{x}+3\sqrt{y}=0\end{cases}}\Leftrightarrow\sqrt{x}=5\sqrt{y}\Leftrightarrow x=25y\)

Khi đó : \(E=\frac{2x+\sqrt{xy}+3y}{x+\sqrt{xy}-y}=\frac{50y+5y+3y}{25y+5y-y}=\frac{58y}{29y}=2\)

(3*x-1)*y+2*căn bậc hai(x)*y+2*x

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.

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

\(\text{méo biết}\)

= căn xy + căn x + căn y còn lại tự tính

Đặt \(\left\{{}\begin{matrix}\sqrt{2x+3}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\)

\(\Rightarrow b\left(b^2+1\right)-3a^2=\left(a^2+1\right)a-3b^2\)

\(\Rightarrow a^3-b^3+3a^2-3b^2+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)\left(3a+3b\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+3a+3b+1\right)=0\)

\(\Leftrightarrow a=b\Rightarrow\sqrt{2x+3}=\sqrt{y}\)

\(\Rightarrow y=2x+3\)

\(\Rightarrow M=x\left(2x+3\right)+3\left(2x+3\right)-4x^2-3\) tới đây chắc chỉ cần bấm máy

B=3,406938828

xcănx=cănx mũ 3

y căn y = căn y mũ 3

Có : \(x-2y-\sqrt{xy}+\sqrt{x}-2\sqrt{y}=0\)

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

\(\Leftrightarrow\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+1\right)=0\)

\(\Leftrightarrow\sqrt{x}=2\sqrt{y}\) (Do \(\sqrt{x}+\sqrt{y}+1>0,\forall x;y>0\))

\(\Leftrightarrow x=4y\)

Khi đó \(P=\dfrac{7y}{\left(2\sqrt{y}+3\sqrt{y}\right).\left(\sqrt{x}+2\sqrt{y}\right)}\)

\(=\dfrac{7y}{5\sqrt{y}.4\sqrt{y}}=\dfrac{7}{20}\)