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a.
\(\left\{{}\begin{matrix}x^3-y^3=16x-4y\\-4=5x^2-y^2\end{matrix}\right.\)
Nhân vế:
\(-4\left(x^3-y^3\right)=\left(16x-4y\right)\left(5x^2-y^2\right)\)
\(\Leftrightarrow21x^3-5x^2y-4xy^2=0\)
\(\Leftrightarrow x\left(7x-4y\right)\left(3x+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4y}{7}\\y=-3x\end{matrix}\right.\)
Thế vào \(y^2=5x^2+4...\)
b. Đề bài không hợp lý ở \(4x^2\)
c.
\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=9\\3x^2+6y^2=3x-12y\end{matrix}\right.\)
Trừ vế:
\(x^3-y^3-3x^2-6y^2=9-3x+12y\)
\(\Leftrightarrow x^3-3x^2+3x-1=y^3+6y^2+12y+8\)
\(\Leftrightarrow\left(x-1\right)^3=\left(y+2\right)^3\)
\(\Leftrightarrow x-1=y+2\)
\(\Leftrightarrow y=x-3\)
Thế vào \(x^2=2y^2=x-4y\) ...
\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}xy\left(x+y\right)=u\\xy+x+y=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}uv-30=0\\u+v-11=0\end{matrix}\right.\) \(\Rightarrow\left(u;v\right)=\left(6;5\right);\left(5;6\right)\)
TH1: \(\left\{{}\begin{matrix}xy\left(x+y\right)=6\\xy+x+y=5\end{matrix}\right.\)
Theo Viet đảo \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)hoặc \(\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\)(vô nghiệm)
TH2: \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\xy=1\end{matrix}\right.\) \(\Rightarrow...\) hoặc \(\left\{{}\begin{matrix}x+y=1\\xy=5\end{matrix}\right.\) (vô nghiệm)
2 câu dưới hình như em hỏi rồi?
a.
\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=16x-4y\\-4=5x^2-y^2\end{matrix}\right.\)
\(\Rightarrow-4\left(x^3-y^3\right)=\left(5x^2-y^2\right)\left(16x-4y\right)\)
\(\Leftrightarrow21x^3-5x^2y-4xy^2=0\)
\(\Leftrightarrow x\left(7x-4y\right)\left(3x+y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\y=\dfrac{7x}{4}\\y=-3x\end{matrix}\right.\)
Lần lượt thế vào \(y^2=5x^2+4\)...
b. Đề bài bất hợp lý, \(4x^2+y^4\) cần là \(4x^4+y^4\)
ý a ở đây bn https://hoc247.net/hoi-dap/toan-10/giai-he-pt-3x-x-2-2-y-2-va-3y-y-2-2-x-2-faq371128.html
b.
Với \(xy=0\) không là nghiệm
Với \(xy\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y^2+1\right)=y\left(5-x^2\right)\\y^2+1=y\left(5-2x\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y^2+1}{y}=\dfrac{5-x^2}{x}\\\dfrac{y^2+1}{y}=5-2x\end{matrix}\right.\)
\(\Rightarrow\dfrac{5-x^2}{x}=5-2x\)
\(\Leftrightarrow5-x^2=5x-2x^2\)
\(\Leftrightarrow...\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+1+y^2=1\\\left(x-1\right)^3+y^3=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2+y^2=1\\\left(x-1\right)^3+y^3=1\end{matrix}\right.\)
Do \(\left(x-1\right)^2+y^2=1\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|\le1\\\left|y\right|\le1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)^3\le\left(x-1\right)^2\\y^3\le y^2\end{matrix}\right.\) \(\Rightarrow\left(x-1\right)^3+y^3\le\left(x-1\right)^2+y^2=1\)
Dấu "=" xảy ra khi và chỉ khi \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=0\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=1\\y=0\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;1\right);\left(2;0\right)\)
Với \(xy=0\) là nghiệm
Với \(xy\ne0\)
\(\Rightarrow\left\{{}\begin{matrix}y-\dfrac{2}{x}+\dfrac{3x}{y}=0\\\dfrac{y}{x}+x+\dfrac{2}{y}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y-\dfrac{2}{x}=-\dfrac{3x}{y}\\x+\dfrac{2}{y}=-\dfrac{y}{x}\end{matrix}\right.\)
\(\Rightarrow\left(y-\dfrac{2}{x}\right)\left(x+\dfrac{2}{y}\right)=3\)
\(\Leftrightarrow xy-\dfrac{4}{xy}-3=0\)
\(\Rightarrow\left(xy\right)^2-3xy-4=0\Rightarrow\left[{}\begin{matrix}xy=-1\\xy=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{y}\\x=\dfrac{4}{y}\end{matrix}\right.\) thế vào \(y^2+x^2y+2x=0\)
\(\Rightarrow\left[{}\begin{matrix}y^2+\dfrac{1}{y}-\dfrac{2}{y}=0\\y^2+\dfrac{16}{y}+\dfrac{8}{y}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y^3=1\\y^3=-24\end{matrix}\right.\)
\(\Leftrightarrow...\)
a, \(\left\{{}\begin{matrix}x+y=4\\\left(x^2+y^2\right)\left(x^3+y^3\right)=280\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left(x^2+y^2\right)\left(x^2+y^2-xy\right)=70\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left(16-2xy\right)\left(16-3xy\right)=70\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\3x^2y^2-40xy+93=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left[{}\begin{matrix}xy=\dfrac{31}{3}\\xy=3\end{matrix}\right.\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=4\\xy=\dfrac{31}{3}\end{matrix}\right.\)
Phương trình này vô nghiệm
Vậy hệ đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(1;3\right);\left(3;1\right)\right\}\)
b, ĐK: \(xy>0\)
\(\left\{{}\begin{matrix}\sqrt{\dfrac{2x}{y}}+\sqrt{\dfrac{2y}{x}}=3\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x}{y}+\dfrac{2y}{x}+4=9\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x^2+y^2\right)=5xy\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(x-2y\right)=0\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}2x=y\\x=2y\end{matrix}\right.\\x-y+xy=3\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}y=2x\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2x\\2x^2-x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2x\\\left(x+1\right)\left(2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=-2\\x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y=3\\x=\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x=2y\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\2y^2+y-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\y=\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
Cộng vế:
\(x^3-y^3+3x^2+3y^2+4x-4y+4=0\)
\(\Leftrightarrow\left(x+1\right)^3-\left(y-1\right)^3+x-y+2=0\)
\(\Leftrightarrow\left(x-y+2\right)\left(x^2+y^2+xy+x-y+2\right)=0\)
\(\Leftrightarrow\left(x-y+2\right)\left[\left(x+\dfrac{y}{2}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2+1\right]=0\)
\(\Leftrightarrow y=x+2\)
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