Đáp án: $x=\dfrac12,y=2$
Giải thích các bước giải:
ĐKXĐ: $y\le\dfrac52, x\le\dfrac34$
Ta có:
$(4x^2+1)x+(y-3)\sqrt{5-2y}=0$
$\to 4x^3+x=-(y-3)\sqrt{5-2y}$
$\to 8x^3+2x=-(2y-6)\sqrt{5-2y}$
$\to 8x^3+2x=-(2y-5-1)\sqrt{5-2y}$
$\to 8x^3+2x=-(2y-5)\sqrt{5-2y}+\sqrt{5-2y}$
$\to (2x)^3+2x=(5-2y)\sqrt{5-2y}+\sqrt{5-2y}$
$\to (2x)^3+2x=(\sqrt{5-2y})^3+\sqrt{5-2y}$
$\to (2x)^3-(\sqrt{5-2y})^3+(2x-\sqrt{5-2y})=0$
$\to (2x-\sqrt{5-2y})(4x^2+2x\sqrt{5-2y}+(5-2y))+(2x-\sqrt{5-2y})=0$
$\to (2x-\sqrt{5-2y})(4x^2+2x\sqrt{5-2y}+(5-2y)+1)=0$
Mà $4x^2+2x\sqrt{5-2y}+(5-2y)+1= (2x+\dfrac12\sqrt{5-2y})^2+\dfrac34(5-2y)+1>0$ do $5-2y\ge 0$
$\to 2x-\sqrt{5-2y}=0$
$\to 2x=\sqrt{5-2y}$
$\to x=\dfrac12\sqrt{5-2y}$
Ta có:
$4x^2+y^2+2\sqrt{3-4x}=7$
$\to 4\cdot (\dfrac12\sqrt{5-2y})^2+y^2+2\sqrt{3-4\cdot \dfrac12\sqrt{5-2y}}=7$
$\to y^2-2y-2+2\sqrt{3-2\sqrt{5-2y}}=0$
$\to (y-1)^2-3+2\sqrt{3-2\sqrt{5-2y}}=0$
Ta có $3-2\sqrt{5-2y}\ge 0\to \dfrac{11}8\le y\le\dfrac52$
Nếu $y>2$
$\to (y-1)^2-3+2\sqrt{3-2\sqrt{5-2y}}>(2-1)^2-3+2\sqrt{3-2\sqrt{5-2\cdot 2}}=0$
$\to$Loại
Nếu $y=2$
$\to (y-1)^2-3+2\sqrt{3-2\sqrt{5-2y}}=(2-1)^2-3+2\sqrt{3-2\sqrt{5-2\cdot 2}}=0$
$\to y=2$ chọn
$\to x=\dfrac12$
Nếu $y<2$
$\to\dfrac{11}8\le y<2$
$\to \dfrac38\le y-1<1$
$\to (y-1)^2<1$
$\to (y-1)^2-3+2\sqrt{3-2\sqrt{5-2y}}<1-3+2\sqrt{3-2\sqrt{5-2\cdot 2}}=0$
$\to y<2$ loại
