Giải hệ phương trình
\(\left\{{}\begin{matrix}x^2+y^2+xy=13\\x^4+y^4+x^2y^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=13\\\left(x^2+y^2\right)^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=13+xy\\\left[\left(x+y\right)^2-2xy\right]^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=13\\\left(13-xy\right)^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy=3\\\left(x+y\right)^2=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) hoặc x+y = -4
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=-4\\xy=3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)
Mọi người có thể giải thích từ dấu tương đương thứ 3 xuống 4. tại sao lại như vậy k?
Giải hệ phương trình sau :
a, \(\left\{{}\begin{matrix}\left(x^2+1\right)\left(y^2+1\right)=10\\\left(x+y\right)\left(xy-1\right)=3\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}x^3-1=2y\\y^3-1=2x\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}2x^2+xy=3x\\2y^2+xy=3y\end{matrix}\right.\)
b) Lấy pt đầu trừ pt dưới thu được:
\(x^3-y^3+2\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2+2\right)=0\)
Do \(x^2+xy+y^2=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+2>0\)
Do đó x = y. Thay vào pt đầu thu được:
\(x^3-2x-1=0\Leftrightarrow\left(x+1\right)\left(x^2-x-1\right)=0\)
c) Lấy pt trên trừ pt dưới:
\(2\left(x^2-y^2\right)-3\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(2x+2y-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\2x+2y-3=0\end{matrix}\right.\)
Auto làm nốt:D
P/s: Is that true?
Giải hệ phương trình:
a)\(\left\{{}\begin{matrix}\left|2x-y\right|-2\left|y-x\right|=1\\3\left|2x-y\right|+\left|x+y\right|=10\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\left(\dfrac{x}{y}\right)^2+\left(\dfrac{x}{y}\right)^3=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\)
Giải hệ pt
a) \(\left\{{}\begin{matrix}x^2+2xy^2=3\\y^3+y+x\left(2xy-1\right)=3\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2+x^3y-xy^2+xy-y=1\\x^4+y^2-xy\left(2x-1\right)=1\end{matrix}\right.\)
Câu a pt đầu là \(x^2+2xy^2=3\) hay \(x^3+2xy^2=3\) vậy nhỉ? Nhìn \(x^2\) chẳng hợp lý chút nào
b. \(\Leftrightarrow\left\{{}\begin{matrix}x^2\left(xy+1\right)-y\left(xy+1\right)+xy+1=2\\\left(x^4+y^2-2x^2y\right)+xy+1=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-y\right)\left(xy+1\right)+xy+1=2\\\left(x^2-y\right)^2+xy+1=2\end{matrix}\right.\)
Trừ vế cho vế:
\(\left(x^2-y\right)\left(xy+1\right)-\left(x^2-y\right)^2=0\)
\(\Leftrightarrow\left(x^2-y\right)\left(xy+1-x^2+y\right)=0\)
\(\Leftrightarrow\left(x^2-y\right)\left[y\left(x+1\right)+\left(x+1\right)\left(1-x\right)\right]=0\)
\(\Leftrightarrow\left(x^2-y\right)\left(x+1\right)\left(y+1-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=x^2\\x=-1\\y=x-1\end{matrix}\right.\)
- Với \(y=x^2\) thế xuống pt dưới:
\(x^4+x^4-x^3\left(2x-1\right)=1\Leftrightarrow x^3=1\Leftrightarrow...\)
....
Hai trường hợp còn lại bạn tự thế tương tự
giải hệ pt :
a,\(\left\{{}\begin{matrix}x^3y\left(1+y\right)+x^2y^2\left(2+y\right)+xy^3-30=0\\x^2y+x\left(1+y+y^2\right)+y-11=0\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}xy^2-2y+3x^2=0\\y^2+x^2y+2x=0\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}3xy+2y=5\\2xy\left(x+y\right)+y^2=5\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
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?
Giải các hệ phương trình sau :
a, \(\left\{{}\begin{matrix}x^2+xy=y^2+1\\3x+y=y^2+3\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}x^2-y^2=4x-2y-3\\x^2+y^2=5\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}x^2+x-xy-2y^2-2y=0\\x^2+y^2=1\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}2\left(y+z\right)=yz\\xy+yz+zx=108\\xyz=180\end{matrix}\right.\)
Giải hệ phương trình \(\left\{{}\begin{matrix}2x^2-y^2-4\left(x-y\right)=1\\x^2\left(x-2\right)^2+2=\left(xy-2y\right)\left(xy-4x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x^2-2x\right)-\left(y^2-4y\right)=1\\\left(x^2-2x\right)^2+2=y\left(x-2\right)x\left(y-4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x^2-2x\right)-\left(y^2-4y\right)=1\\\left(x^2-2x\right)^2+2=\left(x^2-2x\right)\left(y^2-4y\right)\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2-2x=u\\y^2-4y=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2u-v=1\\u^2+2=uv\end{matrix}\right.\) \(\Rightarrow u^2+2=u\left(2u-1\right)\)
\(\Leftrightarrow u^2-u-2=0\Leftrightarrow...\)
giải hệ phương trình:\(\left\{{}\begin{matrix}\left(x-3\right)\left(y+6\right)=xy\\\left(x+2\right)\left(y-2\right)=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy+6x-3y-18=xy\\xy-2x+2y-4=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-3y=18\\-2x+2y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=6\\-x+y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=10\end{matrix}\right.\)
giải hệ phương trình:
a,\(\left\{{}\begin{matrix}\dfrac{1}{2}\left(x+2\right)\left(y+3\right)=\dfrac{1}{2}xy+50\\\dfrac{1}{2}\left(x-2\right)\left(y-2\right)=\dfrac{1}{2}xy-32\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}\dfrac{-1}{2}x+\dfrac{1}{3}y=0\\y-x=1\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}x\left(y-2\right)=\left(x+2\right)\left(y-4\right)\\\left(x-3\right)\left(2y+7\right)=\left(2x-7\right)\left(y+3\right)\end{matrix}\right.\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)\left(y+3\right)=xy+100\\\left(x-2\right)\left(y-2\right)=xy-64\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=94\\-2x-2y=-68\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=26\\y=8\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}-3x+2y=0\\-x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}xy-2x=xy-4x+2y-8\\2xy+7x-6y-21=2xy+6x-7y-21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-2y=-8\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\)
