Giúp mk với!!! Giải hệ phương trình: x^2-xy-2y^2=0 3x+y=1 16/07/2021 Bởi Clara Giúp mk với!!! Giải hệ phương trình: x^2-xy-2y^2=0 3x+y=1
\begin{cases} x^2-xy-2y^2=0\\3x+y=1 \end{cases} `⇔`\begin{cases}(x+y)(x-2y)=0\\3x+y=1 \end{cases} với` x+y=0` `⇒x=-y` `3x+y=3x-x=2x=1` `⇔x=1/2` `⇒y=-1/2` với `x-2y=0` `⇔x=2y` `3x+y=6y+y=7y=1` `⇔y=1/7` `⇔x=2×1/7=2/7` Bình luận
$\begin{array}{l} \left\{ \begin{array}{l} {x^2} – xy – 2{y^2} = 0\\ 3x + y = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {x^2} + xy – 2xy – 2{y^2} = 0\\ 3x + y = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x\left( {x + y} \right) – 2y\left( {x + y} \right) = 0\\ 3x + y = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \left( {x – 2y} \right)\left( {x + y} \right) = 0\\ 3x + y = 1 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x + y = 0\\ 3x + y = 1 \end{array} \right.\\ \left\{ \begin{array}{l} x – 2y = 0\\ 3x + y = 1 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x = \dfrac{1}{2}\\ y = – \dfrac{1}{2} \end{array} \right.\\ \left\{ \begin{array}{l} x = \dfrac{2}{7}\\ y = \dfrac{1}{7} \end{array} \right. \end{array} \right.\\ \Rightarrow \left( {x;y} \right) = \left( {\dfrac{1}{2}; – \dfrac{1}{2}} \right),\left( {\dfrac{2}{7};\dfrac{1}{7}} \right) \end{array}$ Bình luận
\begin{cases} x^2-xy-2y^2=0\\3x+y=1 \end{cases}
`⇔`\begin{cases}(x+y)(x-2y)=0\\3x+y=1 \end{cases}
với` x+y=0`
`⇒x=-y`
`3x+y=3x-x=2x=1`
`⇔x=1/2`
`⇒y=-1/2`
với `x-2y=0`
`⇔x=2y`
`3x+y=6y+y=7y=1`
`⇔y=1/7`
`⇔x=2×1/7=2/7`
$\begin{array}{l} \left\{ \begin{array}{l} {x^2} – xy – 2{y^2} = 0\\ 3x + y = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {x^2} + xy – 2xy – 2{y^2} = 0\\ 3x + y = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x\left( {x + y} \right) – 2y\left( {x + y} \right) = 0\\ 3x + y = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \left( {x – 2y} \right)\left( {x + y} \right) = 0\\ 3x + y = 1 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x + y = 0\\ 3x + y = 1 \end{array} \right.\\ \left\{ \begin{array}{l} x – 2y = 0\\ 3x + y = 1 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x = \dfrac{1}{2}\\ y = – \dfrac{1}{2} \end{array} \right.\\ \left\{ \begin{array}{l} x = \dfrac{2}{7}\\ y = \dfrac{1}{7} \end{array} \right. \end{array} \right.\\ \Rightarrow \left( {x;y} \right) = \left( {\dfrac{1}{2}; – \dfrac{1}{2}} \right),\left( {\dfrac{2}{7};\dfrac{1}{7}} \right) \end{array}$