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a.
⇔ \(\left\{{}\begin{matrix}x-2y=4.3-5\\2x+y=3.3\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}x-2y=7\\2x+y=9\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}-2x+4y=-14\\2x+y=9\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}5y=-5\\2x+y=9\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}y=-1\\2x-1=9\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}y=-1\\x=5\end{matrix}\right.\)
Vậy nghiệm của hpt là: (5;1)
a. Bạn tự giải
b. \(\Leftrightarrow\left\{{}\begin{matrix}x-2y=4m-5\\4x+2y=6m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=4m-5\\5x=10m-5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2m-1\\y=-m+2\end{matrix}\right.\)
\(\dfrac{2}{x}-\dfrac{1}{y}=-1\Rightarrow\dfrac{2}{2m-1}-\dfrac{1}{-m+2}=-1\) (\(m\ne\left\{\dfrac{1}{2};2\right\}\))
\(\Leftrightarrow2\left(-m+2\right)-\left(2m-1\right)=\left(m-2\right)\left(2m-1\right)\)
\(\Leftrightarrow2m^2-m-3=0\Rightarrow\left[{}\begin{matrix}m=-1\\m=\dfrac{3}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-2y=3-m\\4x+2y=6m+12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=m+3\\y=m\end{matrix}\right.\)
\(\Rightarrow x^2+y^2=\left(m+3\right)^2+m^2=2m^2+6m+9=2\left(m+\dfrac{3}{2}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}\)
\(\Rightarrow\left(x^2+y^2\right)_{min}=\dfrac{9}{2}\) khi \(m+\dfrac{3}{2}=0\Rightarrow m=-\dfrac{3}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=m-6\\\left(m+3\right)x-2y=4m-13\end{matrix}\right.\)
Theo điều kiện có nghiệm duy nhất của hệ thì:
\(\frac{m+3}{1}\ne\frac{-2}{-1}\Leftrightarrow m\ne-1\)
Khi đó: \(\left\{{}\begin{matrix}x-y+6=m\\3x-2y+13=4m-mx\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y+6=m\\\frac{3x-2y+13}{4-x}=m\end{matrix}\right.\) \(\Rightarrow x-y+6=\frac{3x-2y+13}{4-x}\)
Đây là biểu thức liên hệ 2 nghiệm ko phụ thuộc m
Muốn chắc chắn hơn, bạn có thể biện luận riêng trường hợp \(x=4\)
a, Ta có ( I ) : \(\left\{{}\begin{matrix}x+y=5\\xy=5\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=5-y\\y\left(5-y\right)=5\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=5-y\\5y-y^2-5=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=5-y\\y^2-5y+5=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=5-y\\y^2-2.\frac{5}{2}y+\left(\frac{5}{2}\right)^2-1,25=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=5-y\\\left(y-2,5\right)^2=1,25\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=5-y\\\left[{}\begin{matrix}y-2,5=\frac{\sqrt{5}}{2}\\y-2,5=-\frac{\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=5-\frac{\sqrt{5}}{2}-2,5=\frac{5-\sqrt{5}}{2}\\x=5-2,5+\frac{\sqrt{5}}{2}=\frac{15-\sqrt{5}}{2}\end{matrix}\right.\\\left[{}\begin{matrix}y=\frac{\sqrt{5}}{2}+2,5\\y=2,5-\frac{\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy hệ phương trình có 2 nghiệm là : \(\left(x,y\right)=\left(\frac{5-\sqrt{5}}{2},\frac{5+\sqrt{5}}{2}\right),\left(\frac{15-\sqrt{5}}{2},\frac{5-\sqrt{5}}{2}\right)\) .
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\3-y+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}x-2x-1=3\\y=2x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\left(-2\right)+1=-3\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}2x+3x-6=4\\y=x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\\ 4,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y+2=3y+8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\\ 5,\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+y}{2}\\\dfrac{3+3y}{2}-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+y}{2}\\3+3y-8y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{y+1}{2}\\y=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{1}{5}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}4x+3x=-6\\\dfrac{x+3y}{3}-\dfrac{y-2}{5}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x=-6\\\dfrac{5\left(x+3y\right)-3\left(y-2\right)}{15}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\5x+15y-3y+6=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\12y=9-5x=9+5\cdot\dfrac{6}{7}=9+\dfrac{30}{7}=\dfrac{93}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\y=\dfrac{93}{7\cdot12}=\dfrac{93}{84}=\dfrac{31}{28}\end{matrix}\right.\)
HPT\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-xy=1-2xy\\\left(x+y\right)\left(1-2xy\right)=x+3y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=1\\x^2+xy=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=1\\y=-\sqrt{2};\sqrt{2}\end{matrix}\right.\)
The vao roi tinh la xong
a) Thay m=1 vào hệ phương trình, ta được:
\(\left\{{}\begin{matrix}x+2y=2\\2x+3y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=4\\2x+3y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=5\\x+2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x+10=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-8\\y=5\end{matrix}\right.\)
Vậy: Khi m=1 thì hệ phương trình có nghiệm duy nhất là (x,y)=(-8;5)
b) Ta có: \(\left\{{}\begin{matrix}x+2y=m+1\\2x+3y=m-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=2m+2\\2x+3y=m-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=m+4\\x+2\cdot\left(m+4\right)=m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2m+8=m+1\\y=m+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-m-7\\y=m+4\end{matrix}\right.\)
Để hệ phương trình có nghiệm (x,y) thỏa mãn x>3 và y<5 thì \(\left\{{}\begin{matrix}-m-7>3\\m+4< 5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-m>10\\m< 1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< -10\\m< 1\end{matrix}\right.\Leftrightarrow m< -10\)
Vậy: Để hệ phương trình có nghiệm (x,y) thỏa mãn x>3 và y<5 thì m<-10