Chủ đề này có 1 trả lời

[luuvanthai]

Giải hệ pt: $\left\{\begin{matrix} 4x^{2}+2xy-2y^{2}+5y-x=3 & \\ 2x^{2}-5xy+2y^{2}+3x=10 & \end{matrix}\right.$

[BaNam]

Giải hệ pt: $\left\{\begin{matrix} 4x^{2}+2xy-2y^{2}+5y-x=3 & \\ 2x^{2}-5xy+2y^{2}+3x=10 & \end{matrix}\right.$

Xét phương trình $^{(1)}$ $\Rightarrow$ $4x^{2}+x(2y-1)-2y^{2}+5y-3=0$ 
$\Leftrightarrow$ $2x(x+y-1)+2(x-y)(x+y-1) +3(x+y-1)=0$

$\Leftrightarrow$ $(x+y-1)(4x-2y+3)=0$

$\Rightarrow$ $\left [\begin{array}{ll} x+y-1=0 \\ 4x-2y+3=0 \end{array}\right.$ 
$+)$ $TH1$: $\Rightarrow$ $\left\{\begin{matrix} x+y=1 \\ 2x^{2}-5xy+2y^{2}+3x=10 \end{matrix}\right.$
$\Leftrightarrow$ $\left\{\begin{matrix} y=1-x \\ (2x-y)(x-2y) +3x=10 \end{matrix}\right.$

$\Leftrightarrow$ $\left\{\begin{matrix} y=1-x \\ (3x-1)(3x-2)+3x-10=0 \end{matrix}\right.$

$\Leftrightarrow$ $\left\{\begin{matrix} y=1-x \\ 9x^{2} -6x-8=0 \end{matrix}\right.$
$\Leftrightarrow$ $\left\{\begin{matrix} y=1-x \\ \left [\begin{array}{ll} x=\frac{4}{3} \\ x=-\frac{2}{3} \end{array}\right. \end{matrix}\right.$
$\Rightarrow$ $\left [\begin{array}{ll} \left\{\begin{matrix} x=\frac{4}{3} \\ y=-\frac{1}{3} \end{matrix}\right. \\ \left\{\begin{matrix} x=-\frac{2}{3} \\ y=\frac{5}{3} \end{matrix}\right. \end{array}\right.$

$+)$ $TH2$: $\Rightarrow$ $\left\{\begin{matrix} 4x-2y+3=0 \\ 2x^{2}-5xy+2y^{2}+3x=10 \end{matrix}\right.$
$\Leftrightarrow$ $\left\{\begin{matrix} 2y=4x+3 \\ 4x^{2}-10xy+4y^{2}+6x=20  \end{matrix}\right.$
$\Leftrightarrow$ $\left\{\begin{matrix} 2y=4x+3 \\ 4x^{2}-5x(4x+3)+(4x+3)^{2}+6x=20  \end{matrix}\right.$
$\Leftrightarrow$ $\left\{\begin{matrix} 2y=4x+3 \\ 15x=11  \end{matrix}\right.$
$\Leftrightarrow$ $\left\{\begin{matrix} y=\frac{89}{30} \\ x=\frac{11}{15}  \end{matrix}\right.$