Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ Bạn tự giải
b/ Hệ tương đương:
\(\left\{{}\begin{matrix}2x+3y=m\\15x-3y=3\end{matrix}\right.\) \(\Rightarrow17x=m+3\Rightarrow x=\frac{m+3}{17}\)
\(\Rightarrow y=5x-1=\frac{5x+15}{17}-1=\frac{5m-2}{17}\)
\(\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{m+3}{17}>0\\\frac{5m-2}{17}>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m>-3\\m>\frac{2}{5}\end{matrix}\right.\) \(\Rightarrow m>\frac{2}{5}\)
a, Thay m=3 vào hpt ta có :
\(\left\{{}\begin{matrix}2x+3y=3\\-5x+y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=3\\-15x+3y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=3\\17x=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{17}\\y=\frac{43}{51}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2x+3y=m\\25x-3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-m=3y\\25x-3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{m-2x}{3}\\25x+3x-m=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3-2x}{3}\\27x=3+m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3+m}{27}\\y=\frac{m-\frac{6+2m}{27}}{3}=\frac{27m-6-2m}{81}\end{matrix}\right.\)
Mà: \(x>0;y< 0\Rightarrow\left\{{}\begin{matrix}\frac{3+m}{27}>0\\\frac{25m-6}{81}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>-3\\m< \frac{6}{25}\end{matrix}\right.\)
\(\Rightarrow m\in\left\{-3;\frac{6}{25}\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}27x=m+3\\25x-3y=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{m+3}{27}\\y=\frac{25x-3}{3}=\frac{25m-6}{81}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x>0\\y< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{m+3}{27}>0\\\frac{25m-6}{81}< 0\end{matrix}\right.\) \(\Rightarrow-3< m< \frac{6}{25}\)
x=\(\dfrac{m-3y}{2}\)
=> \(25.\dfrac{m-3y}{2}-3y=3\)
=> 25(m-3y)-6y=6
=> 25m-75y-6y-6=0
=>25m-81y-6=0
=>25m-6=81y
=>y=\(\dfrac{25m-6}{81}\)
=>x=\(\dfrac{m-1}{27}\)
voi x>0 thi \(\dfrac{m-1}{27}>0\)
=> m-1>0
=> m>1
voi y<0 thi \(\dfrac{25m-6}{81}< 0\)
=> 25m-6<0
=> m<6/25
\(\left\{{}\begin{matrix}6x-3my=-9\\m^2x+3my=4m\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}6x-3my=-9\\\left(m^2+6\right)x=4m-9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{4m-9}{m^2+6}\\y=\frac{3m+8}{m^2+6}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x< 0\\y>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{4m-9}{m^2+6}< 0\\\frac{3m+8}{m^2+6}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m< \frac{9}{4}\\m>-\frac{8}{3}\end{matrix}\right.\)
\(\Rightarrow-\frac{8}{3}< m< \frac{9}{4}\)
mk sẽ hướng dẩn nha.
phần a của 2 câu : tương tự nhé https://hoc24.vn/hoi-dap/question/621828.html
1b) thế \(x=-1;y=3\) --> m
1c) rút x và y theo m rồi thế vào giải
\(\left\{{}\begin{matrix}x+my=9\\mx-3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9-my\\9m-m^2y-3y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=9-my\\y=\dfrac{9m-4}{m^2+3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=9+\dfrac{4m+27}{m^2+3}\\y=\dfrac{9m-4}{m^2+3}\end{matrix}\right.\) --> ...
2b) tương tự rút x và y theo m và biện luận
\(\left\{{}\begin{matrix}3x-my=-9\\mx+2y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{my-9}{3}\\m^2y-9m+6y=48\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{my-9}{3}\\y=\dfrac{9m+48}{m^2+6}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\dfrac{9m^2+48m}{m^2+6}-9}{3}\\y=\dfrac{9m+48}{m^2+6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-18m}{m^2+6}\\y=\dfrac{9m+48}{m^2+6}\end{matrix}\right.\) --> ...
3c) từ \(x+y=7\Rightarrow y=7-x\) thế vào hệ ta được hệ pt 2 ẩn --> m
\(\left\{{}\begin{matrix}2x+3y=m\\-5x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=m\\-15x+3y=-3\end{matrix}\right.\)
\(\Rightarrow17x=m+3\)
\(\Leftrightarrow x=\dfrac{m+3}{17}\)
để x>0 \(\Leftrightarrow\dfrac{m+3}{17}>0\Leftrightarrow m+3>0\Leftrightarrow m>-3\)
còn y> gì bạn cũng làm như zậy nhé :))