Giải HPT $\left \{ {{2x^2+5xy +2y^2 +x+y+1=0} \atop {x^2 + 4xy + y^2 + 12x +12y +10=0}} \right.$

Giải HPT
$\left \{ {{2x^2+5xy +2y^2 +x+y+1=0} \atop {x^2 + 4xy + y^2 + 12x +12y +10=0}} \right.$

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  1. Đáp án:

    `(x;y)\in {(2+\sqrt{41};2-\sqrt{41});(2-\sqrt{41};2+\sqrt{41});({-1+2\sqrt{3}}/3;{-1-2\sqrt{3}}/3);({-1-2\sqrt{3}}/3;{-1+2\sqrt{3}}/3)}`

    Giải thích các bước giải:

    $\quad\begin{cases}2x^2+5xy +2y^2 +x+y+1=0\ (1)\\x^2 + 4xy + y^2 + 12x +12y +10=0\ (2)\end{cases}$

    `<=>`$\begin{cases}4x^2+10xy +4y^2 +2x+2y+2=0\\x^2 + 4xy + y^2 + 12x +12y +10=0\end{cases}$

    `<=>`$\begin{cases}3x^2+6xy +3y^2 -10x-10y-8=0\ (*)\\x^2 + 4xy + y^2 + 12x +12y +10=0\end{cases}$

    (*)`<=>3(x+y)^2-10(x+y)-8=0`

    `<=>3(x+y)^2-12(x+y)+2(x+y)-8=0`

    `<=>3(x+y)(x+y-4)+2(x+y-4)=0`

    `<=>(x+y-4).[3(x+y)+2]=0`

    `<=>`$\left[\begin{array}{l}x+y-4=0\\3(x+y)+2=0\end{array}\right.$

    `<=>`$\left[\begin{array}{l}x+y=4\\x+y=\dfrac{-2}{3}\end{array}\right.$

    $\\$

    +) `TH: x+y=4`

    `=>y=4-x` thay vào `(2)`

    `\qquad x^2+4x(4-x)+(4-x)^2+12x+12(4-x)+10=0`

    `<=>x^2+16x-4x^2+16-8x+x^2+48+10=0`

    `<=>-2x^2+8x+74=0`

    `<=>x^2-4x-37=0`

    `<=>`$\left[\begin{array}{l}x=2+\sqrt{41}\\x=2-\sqrt{41}\end{array}\right.$`=>`$\left[\begin{array}{l}y=4-x=2-\sqrt{41}\\y=4-x=2+\sqrt{41}\end{array}\right.$

    $\\$

    +) `TH: x+y=-2/3`

    `=>y=-x-2/3` thay vào `(2)`

    `\qquad x^2+4x(-x-2/3)+(-x-2/3)^2+12x+12(-x-2/3)+10=0`

    `<=>x^2-4x^2-8/ 3 x+x^2+4/ 3 x+4/ 9 -8+10=0`

    `<=>-2x^2-4/ 3 x+{22}/9=0`

    `<=>x^2+2/3x-{11}/9=0`

    `<=>`$\left[\begin{array}{l}x=\dfrac{-1+2\sqrt{3}}{3}\\x=\dfrac{-1-2\sqrt{3}}{3}\end{array}\right.$

    `=>`$\left[\begin{array}{l}y=-x-\dfrac{2}{3}=\dfrac{-1-2\sqrt{3}}{3}\\y=-x-\dfrac{2}{3}=\dfrac{-1+2\sqrt{3}}{3}\end{array}\right.$

    $\\$

    Vậy hệ phương trình đã cho có nghiệm: 

    `(x;y)\in {(2+\sqrt{41};2-\sqrt{41});(2-\sqrt{41};2+\sqrt{41});({-1+2\sqrt{3}}/3;{-1-2\sqrt{3}}/3);({-1-2\sqrt{3}}/3;{-1+2\sqrt{3}}/3)}`

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