giải hệ phương trình:
1, \(\left\{{}\begin{matrix}x^2\left(1+y^2\right)=2\\1+xy+x^2y^2=3x^2\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\left(x+1\right)^2\left(y+1\right)^2=27xy\\\left(x^2+1\right)\left(y^2+1\right)=10xy\end{matrix}\right.\)
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\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
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?
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
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?
phân tích pt1 thành (x+2)(x2+y2-1)=0
\(\Rightarrow\)x= -2 hoặc y2=1-x2
Nếu x=-2 thay vào pt2
Nếu y2=1-x2.Thay vào pt2 để đưa về biến x
Nhân liên hợp 2 vế vs \(\sqrt{2-x^2}-1\)
a) Ta có: \(\left\{{}\begin{matrix}3x-2\left|y\right|=9\\2x+3\left|y\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-4\left|y\right|=18\\6x+9\left|y\right|=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-13\left|y\right|=15\\3x-2\left|y\right|=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|y\right|=\dfrac{-15}{13}\\3x-2\left|y\right|=9\end{matrix}\right.\Leftrightarrow\)Phương trình vô nghiệmVậy: \(S=\varnothing\)
$\begin{cases}3x-2|y|=9\\2x+3|y|=1\\\end{cases}$
`<=>` $\begin{cases}6x-4|y|=18\\6x+9|y|=3\\\end{cases}$
`<=>` $\begin{cases}13|y|=-15(loại)\\|3x|-2|y|=9\\\end{cases}$
Vậy HPT vô nghiệm
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x^2-2x\right)-\left(y^2-4y\right)=1\\\left(x^2-2x\right)^2+2=y\left(x-2\right)x\left(y-4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x^2-2x\right)-\left(y^2-4y\right)=1\\\left(x^2-2x\right)^2+2=\left(x^2-2x\right)\left(y^2-4y\right)\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2-2x=u\\y^2-4y=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2u-v=1\\u^2+2=uv\end{matrix}\right.\) \(\Rightarrow u^2+2=u\left(2u-1\right)\)
\(\Leftrightarrow u^2-u-2=0\Leftrightarrow...\)
1) \(\left\{{}\begin{matrix}x^2\left(1+y^2\right)=2\\1+xy+x^2y^2=3x^2\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2 +x^2y^2=2\\1+xy+x^2y^2-3x^2=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2=2-x^2y^2\\-3x^2+xy+x^2y^2=-1\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2=2-x^2y^2\\-3\left(2-x^2y^2\right)+xy+x^2y^2=-1\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2=2-x^2y^2\\-6+3x^2y^2+xy+x^2y^2=-1\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2=2-x^2y^2\\-5+4x^2y^2+xy=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2=2-x^2y^2\\-5+4x^2y^2+xy=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2=2-x^2y^2\\\left(4xy+5\right)\left(xy-1\right)=0\end{matrix}\right.\)
Ta có 2 trường hợp:
TH1: \(\left\{{}\begin{matrix}x^2=2-x^2y^2\\4xy+5=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2=2-x^2y^2\\xy=\dfrac{-5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=2-\dfrac{25}{16}\\xy=\dfrac{-5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\dfrac{\sqrt{7}}{4}\\y=\dfrac{-5}{\sqrt{7}}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x^2=2-x^2y^2\\xy=1\end{matrix}\right.\)\(\left\{{}\begin{matrix}x^2=2-1\\xy=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=1\\xy=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Câu 2 mình chưa nghĩ ra. Sorry bạn nheee