[toan 10] giải pt

M

miko_tinhnghich_dangyeu

M

miko_tinhnghich_dangyeu

[tex]\left\{ \begin{array}{l} \sqrt{3x}(1+\frac{1}{x+y}=2 \\\sqrt{7y}(1-\frac{1}{x+y})=4\sqrt{2} \end{array} \right.[/tex]
 
T

tuyn

[TEX]\left{\begin{x\sqrt{y}+2y\sqrt{x}=3x\sqrt{2x-1}}\\{y\sqrt{x}+2x\sqrt{y}=3y\sqrt{2y-1}}[/TEX] \Leftrightarrow [TEX]\left{\begin{x^2y+4xy^2+4xy\sqrt{xy}=9x^2(2x-1)}\\{y^2x+4x^2y+4xy\sqrt{xy}=9y^2(2y-1)}[/TEX]
\Rightarrow [TEX]3xy(y-x)=18(x^3-y^3)-9(x^2-y^2)[/TEX] \Leftrightarrow [TEX](x-y)[18(x^2+xy+y^2)-9(x+y)+3xy]=0[/TEX] \Leftrightarrow [TEX]\left[\begin{x-y=0}\\{18(x^2+xy+y^2)-9(x+y)+3xy=0}[/TEX]
ta có [TEX]A=18(x^2+xy+y^2)-9(x+y)+3xy=9(x^2+y^2)+3xy+9(x+y)^2-9(x+y)[/TEX]
Vì \Leftrightarrow [TEX]x \geq \frac{1}{2}[/TEX] và [TEX]y \geq \frac{1}{2}[/TEX]
\Rightarrow [TEX]x+y \geq 1[/TEX] \Rightarrow [TEX](x+y)^2 \geq x+y[/TEX] \Rightarrow [TEX]A > 0[/TEX]
 
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