Cho hệ pt {2x+my=7 {3x-y=3 Tìm nghiệm của hệ pt theo m 07/10/2021 Bởi Mackenzie Cho hệ pt {2x+my=7 {3x-y=3 Tìm nghiệm của hệ pt theo m
Giải thích các bước giải: $\begin{array}{l}\left\{ \begin{array}{l}2x+my=7\\3x-y=3\end{array} \right.⇔\left\{ \begin{array}{l}6x+3my=21\\6x-2y=6\end{array} \right.\\⇔\left\{ \begin{array}{l}(3m+2)y=15\\3x-y=3\end{array} \right.⇔\left\{ \begin{array}{l}y=\dfrac{15}{3m+2}\\3x-\dfrac{15}{3m+2}=3\end{array} \right.\\⇔\left\{ \begin{array}{l}y=\dfrac{15}{3m+2}\\(3x-3)(3m+2)=15\end{array} \right.⇔\left\{ \begin{array}{l}y=\dfrac{15}{3m+2}\\9mx-9m-6+6x=15\end{array} \right.\\⇔\left\{ \begin{array}{l}y=\dfrac{15}{3m+2}\\(9m+6)x=9m+21\end{array} \right.⇔\left\{ \begin{array}{l}y=\dfrac{15}{3m+2}\\x=\dfrac{9m+21}{9m+6}=\dfrac{3m+7}{3m+2}\end{array} \right.\\\text{Vậy $(x;y)=\left(\dfrac{3m+7}{3m+2};\dfrac{15}{3m+2}\right)$}\end{array}$ Bình luận
$\begin{cases}2x+my=7\\3x-y=3\end{cases}$$⇔\begin{cases}2x+my=7\\3mx-my=3m\end{cases}$ $⇔\begin{cases}2x+3mx=3m+7\\y=3x-3\end{cases}$$⇔\begin{cases}x(3m+2)=3m+7\\y=3x-3\end{cases}$$⇔\begin{cases}x=\dfrac{3m+7}{3m+2}\\y=3.\dfrac{3m+7}{3m+2}-3\end{cases}$$⇔\begin{cases}x=\dfrac{3m+7}{3m+2}\\y=\dfrac{9m+21}{3m+2}-3\end{cases}$$⇔\begin{cases}x=\dfrac{3m+7}{3m+2}\\y=\dfrac{15}{3m+2}\end{cases}$ Vậy hệ phương trình có nghiệm duy nhất `(x;y)=({3m+7}/{3m+2};{15}/{3m+2})` Bình luận
Giải thích các bước giải:
$\begin{array}{l}\left\{ \begin{array}{l}2x+my=7\\3x-y=3\end{array} \right.⇔\left\{ \begin{array}{l}6x+3my=21\\6x-2y=6\end{array} \right.\\⇔\left\{ \begin{array}{l}(3m+2)y=15\\3x-y=3\end{array} \right.⇔\left\{ \begin{array}{l}y=\dfrac{15}{3m+2}\\3x-\dfrac{15}{3m+2}=3\end{array} \right.\\⇔\left\{ \begin{array}{l}y=\dfrac{15}{3m+2}\\(3x-3)(3m+2)=15\end{array} \right.⇔\left\{ \begin{array}{l}y=\dfrac{15}{3m+2}\\9mx-9m-6+6x=15\end{array} \right.\\⇔\left\{ \begin{array}{l}y=\dfrac{15}{3m+2}\\(9m+6)x=9m+21\end{array} \right.⇔\left\{ \begin{array}{l}y=\dfrac{15}{3m+2}\\x=\dfrac{9m+21}{9m+6}=\dfrac{3m+7}{3m+2}\end{array} \right.\\\text{Vậy $(x;y)=\left(\dfrac{3m+7}{3m+2};\dfrac{15}{3m+2}\right)$}\end{array}$
$\begin{cases}2x+my=7\\3x-y=3\end{cases}$
$⇔\begin{cases}2x+my=7\\3mx-my=3m\end{cases}$
$⇔\begin{cases}2x+3mx=3m+7\\y=3x-3\end{cases}$
$⇔\begin{cases}x(3m+2)=3m+7\\y=3x-3\end{cases}$
$⇔\begin{cases}x=\dfrac{3m+7}{3m+2}\\y=3.\dfrac{3m+7}{3m+2}-3\end{cases}$
$⇔\begin{cases}x=\dfrac{3m+7}{3m+2}\\y=\dfrac{9m+21}{3m+2}-3\end{cases}$
$⇔\begin{cases}x=\dfrac{3m+7}{3m+2}\\y=\dfrac{15}{3m+2}\end{cases}$
Vậy hệ phương trình có nghiệm duy nhất `(x;y)=({3m+7}/{3m+2};{15}/{3m+2})`