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mình giải tắt nhé vì mình không giỏi dùng công thức. Thông cảm nha.
1.
\(\left\{{}\begin{matrix}3x-y=2m+3\\x+y=3m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{m}{4}+1\\y=\dfrac{-5m}{4}\end{matrix}\right.\)
vậy phương trình có nghiệm duy nhất là \(\left(\dfrac{m}{4}+1;\dfrac{-5m}{4}\right)\)
Thay vào đẳng thức ta được:
\(\left(\dfrac{m}{4}+1\right)^2+\left(\dfrac{-5m}{4}\right)^2=5\\ \Leftrightarrow x=\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+2y=6m+4\\3x-2y=11-m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=5m+15\\3x-2y=11-m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m+3\\y=2m-1\end{matrix}\right.\)
\(A=x^2-y^2=\left(m+3\right)^2-\left(2m-1\right)^2\)
\(=-3m^2+10m+8=-3\left(m-\frac{5}{3}\right)^2+\frac{49}{3}\le\frac{49}{3}\)
\(A_{max}=\frac{49}{3}\) khi \(m=\frac{5}{3}\)
\(\left\{{}\begin{matrix}mx+y=7\\2x-y=-4\end{matrix}\right.\left(1\right)\)
Ta có: \(2x-y=-4\)
\(\Rightarrow y=2x+4\)
\(P=x^2+y^2=x^2+\left(2x+4\right)^2=x^2+4x^2+16x+16\)
\(P=5x^2+16x+16=5\left(x^2+2.\frac{8}{5}x+\frac{64}{25}\right)+\frac{16}{5}\)
\(P=5\left(x+\frac{8}{5}\right)^2+\frac{16}{5}\)
Do: \(\left(x+\frac{8}{5}\right)^2\ge0\Rightarrow5\left(x+\frac{8}{5}\right)^2+\frac{16}{5}\ge\frac{16}{5}\)
\(P_{Min}=\frac{16}{5}\Leftrightarrow x=-\frac{8}{5}\) Mà: \(y=2x+4\Rightarrow y=\frac{4}{5}\)
Thay \(x,y\) vào phương trình đề cho ta được:
\(m\left(-\frac{8}{5}\right)+\frac{4}{5}=7\)
\(\Leftrightarrow m=-\frac{31}{8}\)
Vậy nếu \(m=-\frac{31}{8}\) thì \(P\) đạt \(Min=\frac{16}{5}\)
ta có : \(\left\{{}\begin{matrix}mx+y=7\\2x-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=7-mx\\2x-7+mx=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=7-mx\\x=\dfrac{11-mx}{2}\end{matrix}\right.\)
\(\Rightarrow P=x^2+y^2=\dfrac{\left(11-mx\right)^2}{4}+\left(7-mx\right)^2\)
\(=\dfrac{121-22mx+m^2x^2}{4}+49-14mx+m^2x^2\)
\(=\dfrac{5m^2x^2-78mx+317}{4}\)
\(=\dfrac{5m^2x^2-2.\sqrt{5}mx+\dfrac{78}{2\sqrt{5}}+\dfrac{1521}{5}+\dfrac{64}{5}}{4}\)
\(=\dfrac{\left(\sqrt{5}mx-\dfrac{78}{2\sqrt{5}}\right)^2+\dfrac{64}{5}}{4}\)
ta có : \(P\) nhỏ nhất khi \(\dfrac{\left(\sqrt{5}mx-\dfrac{78}{2\sqrt{5}}\right)^2+\dfrac{64}{5}}{4}\) nhỏ nhất
\(\Leftrightarrow\left(\sqrt{5}mx-\dfrac{78}{2\sqrt{5}}\right)^2+\dfrac{64}{5}\) nhỏ nhấtta có : \(\left(\sqrt{5}mx-\dfrac{78}{2\sqrt{5}}\right)^2+\dfrac{64}{5}\ge\dfrac{64}{5}\forall mx\)
khi \(\sqrt{5}mx-\dfrac{78}{2\sqrt{5}}=0\Leftrightarrow m=\dfrac{39}{5x}\)
khi đó ta có : \(P=\dfrac{\dfrac{64}{5}}{4}=\dfrac{16}{5}\)
vậy .............................................................................................
\(\left\{{}\begin{matrix}3x-y=2m-1\\x+2y=3m+2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2y=4m-2\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-2y+x+2y=4m-2+3m+2\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x=7m\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\m+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\2y=2m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=m+1\end{matrix}\right.\)
\(x^2+y^2+3\\ =m^2+\left(m+1\right)^2+3\\ =m^2+m^2+2m+1+3\\ =2m^2+2m+4\\ =2\left(m^2+m+2\right)\)
\(=2\left(m^2+m+\dfrac{1}{4}+\dfrac{7}{4}\right)\)
\(=2\left[\left(m+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right]\)
\(=2\left(m+\dfrac{1}{2}\right)^2+\dfrac{7}{2}\ge\dfrac{7}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow m=-\dfrac{1}{2}\)
Vậy ...
