Toán Dùng hệ số bất định tìm m,n 2x² – 3xy + 2y² = m(x-y)² +n(x+ y)² 04/07/2021 By Kennedy Dùng hệ số bất định tìm m,n 2x² – 3xy + 2y² = m(x-y)² +n(x+ y)²
`2x^2 – 3xy + 2y^2 = m(x – y)^2 + n(x + y)^2` `<=> 2x^2 – 3xy + 2y^2 = m(x^2 – 2xy + y^2) + n(x^2 + 2xy + y^2)` `<=> 2x^2 – 3xy + 2y^2 = mx^2 – m. 2xy + my^2 + nx^2 + n. 2xy + ny^2` `<=> 2x^2 – 3xy + 2y^2 = (mx^2 + nx^2) + (-m. 2xy + n. 2xy) + (my^2 + ny^2)` `<=> 2x^2 – 3xy + 2y^2 = (m + n)x^2 – (2m – 2n)xy + (m + n)y^2` `<=>` \(\left\{\begin{matrix}m + n = 2\\2m – 2n = 3\end{matrix}\right.\) `<=>` \(\left\{\begin{matrix}m + n = 2\\2(m – n) = 3\end{matrix}\right.\) `<=>` \(\left\{\begin{matrix}m + n = 2\\m – n = 1,5\end{matrix}\right.\) `<=>` \(\left\{\begin{matrix}m = (2 + 1,5) : 2 = 1,75\\n = 2 – 1,75 = 0,25\end{matrix}\right.\) Vậy `x = 1,75; n = 0,25` Trả lời
Đáp án: $m = \dfrac{7}{4};n = \dfrac{1}{4}$ Giải thích các bước giải: $\begin{array}{l}m{\left( {x – y} \right)^2} + n{\left( {x + y} \right)^2}\\ = m.\left( {{x^2} – 2xy + {y^2}} \right) + n.\left( {{x^2} + 2xy + {y^2}} \right)\\ = \left( {m + n} \right).{x^2} + \left( { – 2m + 2n} \right).xy + \left( {m + n} \right).{y^2}\\ = 2{x^2} – 3xy + 2{y^2}\\ \Leftrightarrow \left\{ \begin{array}{l}m + n = 2\\ – 2m + 2n = – 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m + n = 2\\2m – 2n = 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}2m + 2n = 4\\2m – 2n = 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}4m = 7\\n = 2 – m\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m = \dfrac{7}{4}\\n = 2 – \dfrac{7}{4} = \dfrac{1}{4}\end{array} \right.\\Vậy\,m = \dfrac{7}{4};n = \dfrac{1}{4}\end{array}$ Trả lời
`2x^2 – 3xy + 2y^2 = m(x – y)^2 + n(x + y)^2`
`<=> 2x^2 – 3xy + 2y^2 = m(x^2 – 2xy + y^2) + n(x^2 + 2xy + y^2)`
`<=> 2x^2 – 3xy + 2y^2 = mx^2 – m. 2xy + my^2 + nx^2 + n. 2xy + ny^2`
`<=> 2x^2 – 3xy + 2y^2 = (mx^2 + nx^2) + (-m. 2xy + n. 2xy) + (my^2 + ny^2)`
`<=> 2x^2 – 3xy + 2y^2 = (m + n)x^2 – (2m – 2n)xy + (m + n)y^2`
`<=>` \(\left\{\begin{matrix}m + n = 2\\2m – 2n = 3\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}m + n = 2\\2(m – n) = 3\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}m + n = 2\\m – n = 1,5\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}m = (2 + 1,5) : 2 = 1,75\\n = 2 – 1,75 = 0,25\end{matrix}\right.\)
Vậy `x = 1,75; n = 0,25`
Đáp án: $m = \dfrac{7}{4};n = \dfrac{1}{4}$
Giải thích các bước giải:
$\begin{array}{l}
m{\left( {x – y} \right)^2} + n{\left( {x + y} \right)^2}\\
= m.\left( {{x^2} – 2xy + {y^2}} \right) + n.\left( {{x^2} + 2xy + {y^2}} \right)\\
= \left( {m + n} \right).{x^2} + \left( { – 2m + 2n} \right).xy + \left( {m + n} \right).{y^2}\\
= 2{x^2} – 3xy + 2{y^2}\\
\Leftrightarrow \left\{ \begin{array}{l}
m + n = 2\\
– 2m + 2n = – 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m + n = 2\\
2m – 2n = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2m + 2n = 4\\
2m – 2n = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
4m = 7\\
n = 2 – m
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m = \dfrac{7}{4}\\
n = 2 – \dfrac{7}{4} = \dfrac{1}{4}
\end{array} \right.\\
Vậy\,m = \dfrac{7}{4};n = \dfrac{1}{4}
\end{array}$