\(\left\{{}\begin{matrix}2x^2-xy+3y^2-7x-12y+1=0\\x-y+1=0\end{matrix}\right.\)
Những câu hỏi liên quan
giải hpt:
a) \(\left\{{}\begin{matrix}4x+9y=6\\3x^2+6xy-x+3y=0\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\left(x+y+2\right)\left(2x+2y-1\right)=0\\3x^2-32y^2+5=0\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}2x^2-xy+3y^2=7x+12y-1\\x-y+1=0\end{matrix}\right.\)
Giải HPT
\(\left\{{}\begin{matrix}2x^2-xy+3y^2=7x+12y-1\\x-y+1=0\end{matrix}\right.\)
\(PT\left(2\right)\Leftrightarrow x=y-1\\ PT\left(1\right)\Leftrightarrow2\left(y-1\right)^2+y\left(1-y\right)+3y^2=7\left(y-1\right)+12y-1\\ \Leftrightarrow2y^2-11y+5=0\\ \Leftrightarrow\left[{}\begin{matrix}y=5\Leftrightarrow x=4\\y=\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)
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Giải hệ phương trình \(\left\{{}\begin{matrix}x-y+1=0\\2x^2-xy+3y^2-7x-12y+1=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-y+1=0\\2x^2-xy+3y^2-7x-12y+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\2\left(y-1\right)^2-\left(y-1\right)y+3y^2-7\left(y-1\right)-12y+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\4y^2-22y+10=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\2y^2-11y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\\left(2y^2-10y\right)-\left(y-5\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\2y\left(y-5\right)-\left(y-5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\\left(y-5\right)\left(2y-1\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y-1\\y-5=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=y-1\\2y-1=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=\frac{-1}{2}\\y=\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy hpt đã cho có 2 nghiệm (x,y) \(\in\left\{\left(4;5\right),\left(\frac{-1}{2};\frac{1}{2}\right)\right\}\)
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\(\left\{{}\begin{matrix}x-y+1=0\\2x^2-xy+3y^2-7x-12y+1=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x=y-1\\2\left(y-1\right)^2-\left(y-1\right)y+3y^2-7\left(y-1\right)-12y+1=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x=y-1\\2y^2-4y+2-y^2+y+3y^2-7y+7-12y+1=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x=y-1\\4y^2-22y+10=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\4y^2-20y-2y+10\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x=y-1\\4y\left(y-5\right)-2\left(y-5\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\2\left(y-5\right)\left(2y-1\right)=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x=y-1\\\Leftrightarrow\left[{}\begin{matrix}y-5=0\\2y-1=0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=5-1=4\\x=\frac{1}{2}-1=-\frac{1}{2}\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}y=5\\y=\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là : (a, b) ∈ {4, 5; -1/2, 1/2}
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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?
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a. \(\left\{{}\begin{matrix}x^2-3x+2y=2\\2x^2+y-x=3\end{matrix}\right.\)
b.\(\left\{{}\begin{matrix}x^2+y^2+xy-3y=4\\2x-3y+xy=3\end{matrix}\right.\)
c.\(\left\{{}\begin{matrix}2x^2=y+\frac{1}{y}\\2y^2=x+\frac{1}{x}\end{matrix}\right.\)
d.\(\left\{{}\begin{matrix}x^2-2y^2-xy-2x+7y-3=0\\x^2+y^2-x+y=0\end{matrix}\right.\)
Giải hệ phương trình:
1. left{{}begin{matrix}x+32sqrt{left(3y-xright)left(y+1right)}sqrt{3y-2}-sqrt{dfrac{x+5}{2}}xy-2y-2end{matrix}right.
2. left{{}begin{matrix}sqrt{2y^2-7y+10-xleft(y+3right)}+sqrt{y+1}x+1sqrt{y+1}+dfrac{3}{x+1}x+2yend{matrix}right.
3. left{{}begin{matrix}sqrt{4x-y}-sqrt{3y-4x}12sqrt{3y-4x}+yleft(5x-yright)xleft(4x+yright)-1end{matrix}right.
4. left{{}begin{matrix}9sqrt{dfrac{41}{2}left(x^2+dfrac{1}{2x+y}right)}3+40xx^2+5xy+6y4y^2+9x+9end{matrix}right.
5. left{{}begin{mat...
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Giải hệ phương trình:
1. \(\left\{{}\begin{matrix}x+3=2\sqrt{\left(3y-x\right)\left(y+1\right)}\\\sqrt{3y-2}-\sqrt{\dfrac{x+5}{2}}=xy-2y-2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{2y^2-7y+10-x\left(y+3\right)}+\sqrt{y+1}=x+1\\\sqrt{y+1}+\dfrac{3}{x+1}=x+2y\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}\sqrt{4x-y}-\sqrt{3y-4x}=1\\2\sqrt{3y-4x}+y\left(5x-y\right)=x\left(4x+y\right)-1\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}9\sqrt{\dfrac{41}{2}\left(x^2+\dfrac{1}{2x+y}\right)}=3+40x\\x^2+5xy+6y=4y^2+9x+9\end{matrix}\right.\)
5. \(\left\{{}\begin{matrix}\sqrt{xy+\left(x-y\right)\left(\sqrt{xy}-2\right)}+\sqrt{x}=y+\sqrt{y}\\\left(x+1\right)\left[y+\sqrt{xy}+x\left(1-x\right)\right]=4\end{matrix}\right.\)
6. \(\left\{{}\begin{matrix}x^4-x^3+3x^2-4y-1=0\\\sqrt{\dfrac{x^2+4y^2}{2}}+\sqrt{\dfrac{x^2+2xy+4y^2}{3}}=x+2y\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}x^3-12z^2+48z-64=0\\y^3-12x^2+48x-64=0\\z^3-12y^2+48y-64=0\end{matrix}\right.\)
15) \(\left\{{}\begin{matrix}3x+2y=7\\x^2+y^2-7x+xy=0\end{matrix}\right.\)
16)\(\left\{{}\begin{matrix}2x+3y=5\\x^2+xy+y^2-4x=-1\end{matrix}\right.\)
>< giúp với ạ
\(\left\{{}\begin{matrix}3x+2y=7\\x^2+y^2-7x+xy=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2y=7\\x^2+y^2-3x^2-2xy+xy=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=7\\-2x^2-xy+y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=7\\-\left(x+y\right)\left(2x-y\right)=0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}3x+2y=7\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=-7\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}3x+2y=7\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy .......
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giải hệ pt :
a, \(\left\{{}\begin{matrix}3y=\dfrac{y^2+2}{x^2}\\3x=\dfrac{x^2+2}{y^2}\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}x^2y+xy^2+x-5y=0\\2xy+y^2-5y+1=0\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}x^2+y^2+xy+2y+x=2\\2x^2-y^2-2y-2=0\end{matrix}\right.\)
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
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b.
Với \(xy=0\) không là nghiệm
Với \(xy\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y^2+1\right)=y\left(5-x^2\right)\\y^2+1=y\left(5-2x\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y^2+1}{y}=\dfrac{5-x^2}{x}\\\dfrac{y^2+1}{y}=5-2x\end{matrix}\right.\)
\(\Rightarrow\dfrac{5-x^2}{x}=5-2x\)
\(\Leftrightarrow5-x^2=5x-2x^2\)
\(\Leftrightarrow...\)
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c.
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x\left(y+1\right)+\left(y+1\right)^2=3\\2x^2-\left(y+1\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x\left(y+1\right)+\left(y+1\right)^2=3\\6x^2-3\left(y+1\right)^2=3\end{matrix}\right.\)
\(\Rightarrow5x^2-x\left(y+1\right)-4\left(y+1\right)^2=0\)
\(\Leftrightarrow\left(x-y-1\right)\left(5x+4\left(y+1\right)\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=x-1\\y=-\dfrac{5x+4}{4}\end{matrix}\right.\)
Thế vào 1 trong 2 pt ban đầu...
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Giải phương trình:
1. \(\left\{{}\begin{matrix}5x-2y=-9\\4x+3y=2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}2x+y-4=0\\x+2y-5=0\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}2x+3y-7=0\\x+2y-4=0\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}5x+6y=17\\9x-y=7\end{matrix}\right.\)
1)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=-27\\8x+6y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2y=5x+9\\23x=-23\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(-1;2\right)\)
2)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x+y=4\\2x+4y=10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3y=-6\\x=5-2y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;2\right)\)
3)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=14\\3x+6y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=4-x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(2;1\right)\)
4)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}5x+6y=17\\54x-6y=42\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}59x=59\\y=9x-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;2\right)\)
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