Giải HPT $\left \{ {{2x^2+5xy +2y^2 +x+y+1=0} \atop {x^2 + 4xy + y^2 + 12x +12y +10=0}} \right.$ 19/07/2021 Bởi Arianna Giải HPT $\left \{ {{2x^2+5xy +2y^2 +x+y+1=0} \atop {x^2 + 4xy + y^2 + 12x +12y +10=0}} \right.$
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)}` Bình luận
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)}`