(2x-1).(4y+2)=-30 Tìm a,b€z biết a.b=18 và a+b =11 05/12/2021 Bởi Katherine (2x-1).(4y+2)=-30 Tìm a,b€z biết a.b=18 và a+b =11
$a) (2x-1).(4y+2)=-30$ Ta có: 1) \(\left[ \begin{array}{l}2x-1=1\\4y+2=-30\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=1\\y=-8\end{array} \right.\) 2) \(\left[ \begin{array}{l}2x-1=2\\4y+2=-15\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=\dfrac{3}{2}\\y=\dfrac{-13}{4}\end{array}(vl) \right.\) 3) \(\left[ \begin{array}{l}2x-1=3\\4y+2=-10\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=2\\y=-3\end{array} \right.\) b) Ta có: $\begin{cases}ab=18\\a+b=11\end{cases}$ $⇔ \begin{cases}a=9\\b=2\end{cases}$ Bình luận
$a) (2x-1).(4y+2)=-30$
Ta có:
1) \(\left[ \begin{array}{l}2x-1=1\\4y+2=-30\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=1\\y=-8\end{array} \right.\)
2) \(\left[ \begin{array}{l}2x-1=2\\4y+2=-15\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=\dfrac{3}{2}\\y=\dfrac{-13}{4}\end{array}(vl) \right.\)
3) \(\left[ \begin{array}{l}2x-1=3\\4y+2=-10\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=2\\y=-3\end{array} \right.\)
b) Ta có: $\begin{cases}ab=18\\a+b=11\end{cases}$
$⇔ \begin{cases}a=9\\b=2\end{cases}$