/x+y/ +/y-2/=2
/x-1/+y =3
giai he phuong trinh
Những câu hỏi liên quan
cho he phuong trinh 3x-y=2m+3 va x+2y=3m+1 tim m de he phuong trinh co 2 nghiem x y thoa man x^2+y^2=5
\(\hept{\begin{cases}3x-y=2m+3\\x+2y=3m+1\end{cases}}\Leftrightarrow\hept{\begin{cases}6x-2y=4m+6\\x+2y=3m+1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=m+1\\y=m\end{cases}}\)khi đó: \(^{x^2+y^2=5\Leftrightarrow2m^2+2m+1=5\Leftrightarrow2m^2+2m-4=0\Leftrightarrow\orbr{\begin{cases}m=1\\m=-2\end{cases}}}\)
giai he phuong trinh
x+2\x+1\y=4
1\x^2+1\xy+x\y=3
giai he phuong trinh x/x-1 + 2y/y+2 = 3 va 2x/x-1 - y/y+2 = -4
Cho he phuong trinh
\(\left(a+1\right)x-ay=5\) (1)
\(x+ay=a^2+4a\) (2)
Tim gia tri cua a thuoc Z sao cho he phuong trinh co nghiem (x;y) voi x, y thuoc Z
Lấy (1) cộng (2), ta có:
\(\left(2a+1\right)x=a^2+4a+5\)\(\Rightarrow x=\dfrac{a^2+4a+5}{2a+1}\)
Thay vào (1): \(\dfrac{\left(a^2+4a+5\right)\left(a+1\right)-10a-5}{2a+1}.\dfrac{1}{a}\)\(=\dfrac{a^3+5a^2-a}{2a+1}.\dfrac{1}{a}=\dfrac{a^2+5a-1}{2a+1}\)
Để x,y nguyên thì \(\left\{{}\begin{matrix}a^2+4a+5⋮2a+1\\a^2+5a-1⋮2a+1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\left(a+2\right)+2a+5⋮2a+1\\a^2+2a+3a-1⋮2a+1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}4⋮2a+1\\a+2⋮2a+1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}4⋮2a+1\\3⋮2a+1\end{matrix}\right.\)\(\Rightarrow2a+1\in\left\{\pm1\right\}\)\(\Rightarrow a\in\left\{-1;0\right\}\)
Vậy với a=-1;0 thì hpt có nghiệm (x;y) với x,y thuộc Z.
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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 \(m+\dfrac{3}{2}=0\Rightarrow m=-\dfrac{3}{2}\)
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giai he phuong trinh
2(x+y) + √(x+1) = 4
(x+y) - 3√(x+1)=-5
\(-\int^{2\left(x+y\right)+\sqrt{x+1}=4}_{2\left(x+y\right)-6\sqrt{x+1}=-10}\Leftrightarrow\int^{7\sqrt{x+1}=14}_{x+y-3\sqrt{x+1}=-5}\Leftrightarrow\int^{\sqrt{x+1}=2}_{x+y-6=-5}\Leftrightarrow\int^{x=3}_{y=-2}\) => vậy..
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giai he phuong trinh sau :
x^3 - x^2 y^2 - y^3 + 1 = 0 va x^3 + xy - 2 = 0
tim m de he phuong trinh va phuong trinh co nghiem
\(a,\sqrt{x^2+3x+2m}=\sqrt{4x-x^2}\)
b, \(\left\{{}\begin{matrix}x+y+1=x\\x^2+y^2=m\end{matrix}\right.\)
giai he phuong trinh\(\left\{{}\begin{matrix}x^3+y^3=1\\x^5+y^5=x^2+y^2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^3+y^3=^{ }1\left(1\right)\\x^5+y^5=x^2+y^2\left(2\right)\end{matrix}\right.\)
(2)\(\Leftrightarrow x^5-x^2+y^5-y^2=0\)
\(\Leftrightarrow x^2\left(x^3-1\right)+y^2\left(y^3-1\right)=0\)
\(\Leftrightarrow x^2\left(-y\right)^3+y^2\left(-x\right)^3=0\)
\(\Leftrightarrow x^2y^3+y^2x^3=0\)
\(\Leftrightarrow x^2y^2\left(x+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\Rightarrow y=1\\y=0\Rightarrow x=1\\x=-y\left(loại\right)\end{matrix}\right.\)
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