cho he phuong trinh:
\(\left\{{}\begin{matrix}x+2y=m+1\\2x+3y=m-2\end{matrix}\right.\)
a. Giai he pt vs m=1
b. Tim m de he pt co nghiem (x;y) thoa man \(\left\{{}\begin{matrix}x>3\\y< 5\end{matrix}\right.\)
cho he phuong trinh \(\left\{{}\begin{matrix}x-y-m+6=0\\\left(m+3\right)x-2y-4m+13=0\end{matrix}\right.\)
Tim m de he phuong trinh co nghiem duy nhat, voi dieu kien do, tim he thuc lien he giua x va y khong phu thuoc vao m
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=m-6\\\left(m+3\right)x-2y=4m-13\end{matrix}\right.\)
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 \(x=4\)
Cho he phuong trinh: \(\left\{{}\begin{matrix}x-2y=3-m\\2x+y=3.\left(m+2\right)\end{matrix}\right.\)
Goi (x;y) la nghiem cua he phuong trinh. Tim m de \(x^2+y^2\) dat GTNN
\(\left\{{}\begin{matrix}x-2y=3-m\\4x+2y=6m+12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=m+3\\y=m\end{matrix}\right.\)
\(\Rightarrow x^2+y^2=\left(m+3\right)^2+m^2=2m^2+6m+9=2\left(m+\dfrac{3}{2}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}\)
\(\Rightarrow\left(x^2+y^2\right)_{min}=\dfrac{9}{2}\) khi \(m+\dfrac{3}{2}=0\Rightarrow m=-\dfrac{3}{2}\)
Bai 2: cho hpt \(\left\{{}\begin{matrix}x-2y=4m-5\\2x+y=3m\end{matrix}\right.\)
a) giai pt khi m=3
b) Tim de pt co nghiem (x,y) thoa man 2x−1y=−12x−1y=−1
(mink dag can gap)
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)
tim m de he pt co 5 nghiem
\(\left\{{}\begin{matrix}x^3-mx=y\\y^3-my=x\end{matrix}\right.\)
Bai 2: cho hpt\(\left\{{}\begin{matrix}x-2y=4m-5\\2x+y=3m\end{matrix}\right.\)
a) giai pt khi m=3
b) Tim de pt co nghiem (x,y) thoa man \(\dfrac{2}{x}-\dfrac{1}{y}=-1\)
(mink dag can gap)
a. Bạn tự giải
b. \(\Leftrightarrow\left\{{}\begin{matrix}x-2y=4m-5\\4x+2y=6m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=4m-5\\5x=10m-5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2m-1\\y=-m+2\end{matrix}\right.\)
\(\dfrac{2}{x}-\dfrac{1}{y}=-1\Rightarrow\dfrac{2}{2m-1}-\dfrac{1}{-m+2}=-1\) (\(m\ne\left\{\dfrac{1}{2};2\right\}\))
\(\Leftrightarrow2\left(-m+2\right)-\left(2m-1\right)=\left(m-2\right)\left(2m-1\right)\)
\(\Leftrightarrow2m^2-m-3=0\Rightarrow\left[{}\begin{matrix}m=-1\\m=\dfrac{3}{2}\end{matrix}\right.\)
tim m de he phuong trinh va phuong trinh co nghiem
\(a,\sqrt{x^2+3x+2m}=\sqrt{4x-x^2}\)
b, \(\left\{{}\begin{matrix}x+y+1=x\\x^2+y^2=m\end{matrix}\right.\)
cho he ptrinh
\(\left\{{}\begin{matrix}mx+2my=m+1\\x+\left(m+1\right)y=2\end{matrix}\right.\)
a,giai he khi m=\(\sqrt{2}\)
b,tim m de he co nghiem duy nhat \(\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\)
tim m de pt co nghiem duy nhat :\(\left\{{}\begin{matrix}x-2y=\dfrac{my}{x}\\y-2x=\dfrac{mx}{y}\end{matrix}\right.\)
ĐKXĐ: \(xy\ne0\)
- Với \(m=0\Rightarrow x=y=0\) (ktm ĐKXĐ) \(\Rightarrow\) hpt vô nghiệm (ktm)
- Với \(m\ne0\)
\(\Rightarrow\left\{{}\begin{matrix}x^2-2xy=my\\y^2-2xy=mx\end{matrix}\right.\)
\(\Rightarrow x^2-y^2=m\left(y-x\right)\)
\(\Rightarrow\left(x-y\right)\left(x+y+m\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y=x\\y=-x-m\end{matrix}\right.\)
- Với \(y=x\Rightarrow-x=m\Rightarrow x=y=-m\)
- Với \(y=-x-m\)
\(\Rightarrow x^2-2x\left(-x-m\right)=m\left(-x-m\right)\)
\(\Rightarrow3x^2+3mx+m^2=0\)
\(\Delta=9m^2-12m^2=-3m^2< 0\Rightarrow\) luôn vô nghiệm với \(m\ne0\)
Vậy với \(m\ne0\) hệ có nghiệm duy nhất \(x=y=-m\) (thỏa mãn)
\(\Rightarrow m\ne0\)
Cho he phuong trinh sau:
\(\hept{\begin{cases}\left(m+1\right)x+my=2m-1\\mx-y=m^2-2\end{cases}}\)
Tim m de he phuong trinh co nghiem duy nhat (x;y) thoa man P= xy dat gia tri lon nhat.