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\(\left\{{}\begin{matrix}3x-y=2m-1\\x+2y=3m+2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2y=4m-2\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-2y+x+2y=4m-2+3m+2\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x=7m\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\m+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\2y=2m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=m+1\end{matrix}\right.\)
\(x^2+y^2+3\\ =m^2+\left(m+1\right)^2+3\\ =m^2+m^2+2m+1+3\\ =2m^2+2m+4\\ =2\left(m^2+m+2\right)\)
\(=2\left(m^2+m+\dfrac{1}{4}+\dfrac{7}{4}\right)\)
\(=2\left[\left(m+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right]\)
\(=2\left(m+\dfrac{1}{2}\right)^2+\dfrac{7}{2}\ge\dfrac{7}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow m=-\dfrac{1}{2}\)
Vậy ...
\(\Leftrightarrow\left\{{}\begin{matrix}2x+2y=6m+4\\3x-2y=11-m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=5m+15\\3x-2y=11-m\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=m+3\\y=2m-1\end{matrix}\right.\)
\(A=x^2-y^2=\left(m+3\right)^2-\left(2m-1\right)^2\)
\(=-3m^2+10m+8=-3\left(m-\frac{5}{3}\right)^2+\frac{49}{3}\le\frac{49}{3}\)
\(A_{max}=\frac{49}{3}\) khi \(m=\frac{5}{3}\)
HPT\(\Leftrightarrow\hept{\begin{cases}x=3m+2-y\\3\left(3m+2-y\right)-2y+m-11=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=3m+2-y\\-5y+10m-5=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=3m+2-\left(2m-1\right)\\y=2m-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=m+3\\y=2m-1\end{cases}}\)
Ta co:
\(x^2-y^2=\left(m+3\right)^2-\left(2m-1\right)^2=-3m^2+10m+8=-3\left(m-\frac{5}{3}\right)^2+\frac{49}{3}\le\frac{49}{3}\)
Dau '=' xay ra khi \(m=\frac{5}{3}\)
\(\Rightarrow\left(x;y\right)=\left(\frac{14}{3};\frac{7}{3}\right)\)
Vay cap nghiem (x;y) de \(x^2-y^2\)dat max la \(\left(\frac{14}{3};\frac{7}{3}\right)\)
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 \(m=\frac{5}{3}\)