Toán 9 HPT $\left\{\begin{matrix} (2x+4y-1)\sqrt{2x-y-1}=(4x-2y-3)\sqrt{x+2y} \\ ...\end{matrix}\right.$

Hằng Suk ! said:

[tex]\left\{\begin{matrix} {\color{Red} {(2x+4-1)}}\sqrt{2x-y-1}=(4x-2y-3)\sqrt{x+2y} \\ x^{2}+8x+5-2(3y+2)\sqrt{4x-3y} =2\sqrt{2x^{2}+5x+2} \end{matrix}\right.[/tex]
help me....

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Xem lại đề $PT(1)$ chỗ màu đỏ.
Rất hiếm khi lại ra đề như vậy.

[tex]\left\{\begin{matrix} (2x+4y-1)\sqrt{2x-y-1}=(4x-2y-3)\sqrt{x+2y}(1) \\ x^{2}+8x+5-2(3y+2)\sqrt{4x-3y} =2\sqrt{2x^{2}+5x+2}(2) \end{matrix}\right.[/tex]
ĐKXĐ:...
[tex]PT(1)\Leftrightarrow (2x+4y)\sqrt{2x-y-1}-\sqrt{2x-y-1}=(4x-2y-2)\sqrt{x+2y}-\sqrt{x+2y}\\\Leftrightarrow 2(x+2y)\sqrt{2x-y-1}-\sqrt{2x-y-1}=2(2x-y-1)\sqrt{x+2y}-\sqrt{x+2y}\\\Leftrightarrow 2[(x+2y)\sqrt{2x-y-1}-(2x-y-1)\sqrt{x+2y}]=\sqrt{2x-y-1}-\sqrt{x+2y}\\\Leftrightarrow \left ( \sqrt{x+2y}-\sqrt{2x-y-1} \right )\left ( 2\sqrt{(x+2y)(2x-y-1)}+1 \right )=0\\\Leftrightarrow \sqrt{x+2y}-\sqrt{2x-y-1}=0\\\Leftrightarrow \sqrt{x+2y}=\sqrt{2x-y-1}\\\Rightarrow x+2y=2x-y-1\\\Leftrightarrow x=3y+1[/tex]
Thay vào [TEX]PT(2)[/TEX] ta được:
[tex]x^{2}+8x+5-2(x+1)\sqrt{3x+1} =2\sqrt{2x^{2}+5x+2}\\\Leftrightarrow \left [ (x^2+2x+1)- 2(x+1)\sqrt{3x+1}+(3x+1)\right ]+\left [ (x+2)-2\sqrt{2x^2+5x+2}+(2x+1) \right ]=0\\\Leftrightarrow \left [ (x+1)-\sqrt{3x+1} \right ]^2+\left [ \sqrt{x+1}-\sqrt{2x-1} \right ]^2=0\\\Leftrightarrow \left\{\begin{matrix} (x+1)-\sqrt{3x+1} =0\\ \sqrt{x+1}-\sqrt{2x-1} =0 \end{matrix}\right.\\\Leftrightarrow x=...\Rightarrow y=...[/tex]