\begin{cases}(2x+3)\sqrt{4x-1}+(2y+3)\sqrt{4y-1}=2\sqrt{(2x+3)(2y+3)} \\ x+y=4xy \end{cases}
quên mất k để ý:sau khi biến đổi hệ pt:\begin{cases}(\sqrt{(2x+3)\sqrt{\frac{x}{y}}}-\sqrt{(2y+3)\sqrt{\frac{y}{x}}})^2=0 \\ x+y=4xy \end{cases}

Cách lm hơi ngu nên thông cảm:
Xét$ x=0,y=0$ k phải là no của hpt.
Xét $x,y\neq0,Ta có:\frac{x}{y}=4x-1;\frac{y}{x}=4y-1$
$hpt\Leftrightarrow \begin{cases}2(x\sqrt{\frac{x}{y}}+y\sqrt{\frac{y}{x}})+3(\sqrt{x/y}+\sqrt{y/x})=2\sqrt{4xy+6(x+y)+9} \\ x+y=4xy \end{cases}$
Đặt $$:S=x+y;P=xy\Rightarrow hpt\Leftrightarrow \begin{cases}2\frac{S^2-2P}{\sqrt{P}}+3\frac{S}{\sqrt{P}}=2\sqrt{4P+6S+9} \\ S=4P \end{cases}\Leftrightarrow \begin{cases}2\frac{16P^2-2P}{\sqrt{P}}+3\frac{4P}{\sqrt{P}}=2\sqrt{4P+24P+9} \\ S= 4P\end{cases}\Leftrightarrow \begin{cases}8\sqrt{P}(4P+1)-2\sqrt{28P+9}=0 \\ S=4P \end{cases}\begin{cases}P=\frac{1}{4} \\ S=1 \end{cases}$$
Với $S=1;P=1/4\Rightarrow x,y là no của pt:t^2-t+\frac{1}{4}=0\Leftrightarrow \frac{1}{4}(2t-1)^2=0\Leftrightarrow \begin{cases}x=1/2 \\ y=1/2 \end{cases}$


tick để đáp án đúng đi e –  ๖ۣۜTQT☾♋☽ 15-12-16 05:54 AM
hihi thank a –  ๖ۣۜTõn♥ 15-12-16 05:53 AM
gãy tay mất...............đúng tick v nhé e –  ๖ۣۜTQT☾♋☽ 14-12-16 04:31 AM

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