Giải hệ phương trình: \(\left\{{}\begin{matrix}x^3+3xy^2=6xy-3x-49\\x^2-6xy+y^2=10y-25x-9\end{matrix}\right.\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 3 2021
Từ pt dưới:
\(x^2+9y^2=6xy\Leftrightarrow x^2-6xy+9y^2=0\)
\(\Leftrightarrow\left(x-3y\right)^2=0\Leftrightarrow x-3y=0\Leftrightarrow x=3y\)
Thế lên pt trên: \(2.\left(3y\right)^2+y^2=19\)
\(\Leftrightarrow19y^2=19\Leftrightarrow y^2=1\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=3\\y=-1\Rightarrow x=-3\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 3 2021
\(8x^3-12x^2y+6xy^2-y^3=8\)
\(\Leftrightarrow\left(2x-y\right)^3=8\)
\(\Leftrightarrow2x-y=2\)
\(\Rightarrow y=2x-2\)
Thế xuống pt dưới:
\(\left(x^2-2x-2\right)\left(-3x^2+6x-9\right)=14\)
Đặt \(x^2-2x=t\)
\(\Rightarrow\left(t-2\right)\left(-3t-9\right)=14\)
\(\Leftrightarrow...\)
10 tháng 2 2022
\(\left\{{}\begin{matrix}9x^2-3xy+2y^2=23\\7x^2+6xy-8y^2=-37\end{matrix}\right.\)\(\left(hpt\right)\)
\(đặt:x=t.y\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}9\left(t.y\right)^2-3t.y^2+2y^2=23\left(1\right)\\7\left(ty\right)^2+6t.y^2-8y^2=-37\left(2\right)\end{matrix}\right.\)
\(\Rightarrow-37\left[9\left(t.y\right)^2-3ty^2+2y^2\right]=23\left[7\left(ty\right)^2+6ty^2-8y^2\right]\)
\(\Leftrightarrow494\left(ty\right)^2+27ty^2-110y^2=0\left(3\right)\)
\(x=y=0\) \(không\) \(là\) \(nghiệm\) \(hpt\)
\(y\ne0\Rightarrow\left(3\right)\Leftrightarrow494t^2+27t-110=0\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{110}{247}\Rightarrow x=\dfrac{110}{247}.y\left(4\right)\\t=-\dfrac{1}{2}\Rightarrow x=-\dfrac{1}{2}.y\left(5\right)\end{matrix}\right.\)
\(thay\left(4\right)và\left(5\right)vào-hpt\Rightarrow x,y=.....\)(đến đây dễ rồi bạn tự tìm x,y)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 1 2021
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-2x+2y^2-y-1=0\\2y^2+2x+y+1-6xy=0\end{matrix}\right.\)
Cộng vế với vế:
\(2x^2+4y^2-6xy=0\)
\(\Leftrightarrow\left(x-y\right)\left(x-2y\right)=0\)
Thế vào 1 trong 2 pt ban đầu
24 tháng 11 2023
b: \(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\\x-3y-3=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\\x=3y+3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(3y+3\right)^2+y^2-2\left(3y+3\right)-2y-23=0\\x=3y+3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}9y^2+18y+9+y^2-6y-6-2y-23=0\\x=3y+3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}10y^2+10y-20=0\\x=3y+3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y^2+y-2=0\\x=3y+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(y+2\right)\left(y-1\right)=0\\x=3y+3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\in\left\{-2;1\right\}\\x=3y+3\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in\left\{\left(-3;-2\right);\left(6;1\right)\right\}\)
a: \(\left\{{}\begin{matrix}3x^2+6xy-x+3y=0\\4x-9y=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}9y=4x-6\\3x^2+6xy-x+3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{9}x-\dfrac{2}{3}\\3x^2+6x\cdot\left(\dfrac{4}{9}x-\dfrac{2}{3}\right)-x+3\cdot\left(\dfrac{4}{9}x-\dfrac{2}{3}\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x^2+\dfrac{8}{3}x^2-4x-x+\dfrac{4}{3}x-2=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{17}{3}x^2-\dfrac{11}{3}x-2=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}17x^2-11x-6=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x-1\right)\left(17x+6\right)=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}17x+6=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=\dfrac{4}{9}\cdot1-\dfrac{2}{3}=\dfrac{4}{9}-\dfrac{2}{3}=-\dfrac{2}{9}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{6}{17}\\y=\dfrac{4}{9}\cdot\dfrac{-6}{17}-\dfrac{2}{3}=\dfrac{-14}{17}\end{matrix}\right.\end{matrix}\right.\)