Giải hệ phương trình
1. \(\left\{{}\begin{matrix}x^2+y^2+2x+2y=\left(x+2\right)\left(y+2\right)\\\left(\frac{x}{y+2}\right)^2+\left(\frac{y}{x+2}\right)^2=1\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}x^2-2xy-6=6y+2x\\\frac{3x^2}{y+1}=4-x\end{matrix}\right.\)
3.\(\left\{{}\begin{matrix}x^2-y=y^2-x\\x^2-x=y+3\end{matrix}\right.\)
4.\(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=\frac{9}{2}\\xy+\frac{1}{xy}+\frac{x}{y}+\frac{y}{x}=5\end{matrix}\right.\)
6.\(\left\{{}\begin{matrix}x^3\left(x-y\right)+x^2y^2=1\\x^2\left(xy+3\right)-3xy=3\end{matrix}\right.\)
7.\(\left\{{}\begin{matrix}x^2+3y-6x=0\\9x^2-6xy^2+y^4-3y+9=0\end{matrix}\right.\)
8.\(\left\{{}\begin{matrix}x^2+y^2+xy=1\\x+y-xy=2y^2-x^2\end{matrix}\right.\)
9.\(\left\{{}\begin{matrix}8x^3-y=y^3-2x\\x^2+y^2=x+2y\end{matrix}\right.\)
10.\(\left\{{}\begin{matrix}2x^2-3xy+y^2+x-y=0\\x^2+x+1=y^2\end{matrix}\right.\)
11.\(\left\{{}\begin{matrix}\left(x^2+y^2\right)\left(x+y+2\right)=4\left(y+2\right)\\x^2+y^2+\left(y+2\right)\left(x+y+2\right)=4\left(y+2\right)\end{matrix}\right.\)
12. \(\left\{{}\begin{matrix}x^2+7=4y^2+4y\\x^2+3xy+2y^2+x+y=0\end{matrix}\right.\)
13. \(\left\{{}\begin{matrix}x^2+y^2=5\\x^3+2y^3+\left(x-5\right)^2+\left(y+5\right)^2=55\end{matrix}\right.\)
14. \(\left\{{}\begin{matrix}\frac{1}{x^2}+\frac{1}{y^2}=3+x^2y^2\\\frac{1}{x^3}+\frac{1}{y^3}+3=x^3y^3\end{matrix}\right.\)
15.\(\left\{{}\begin{matrix}x^2+y^2+4x+2y=3\\x^2+7y^2-4xy+6y=13\end{matrix}\right.\)
16. \(\left\{{}\begin{matrix}x^2-5xy+x-5y^2=42\\7xy+6y^2+42=x\end{matrix}\right.\)
17.\(\left\{{}\begin{matrix}x^2+xy+y^2=13\\x^4+x^2y^2+y^4=91\end{matrix}\right.\)
18.\(\left\{{}\begin{matrix}x^2=\left(2-y\right)\left(2+y\right)\\2x^3=\left(x+y\right)\left(4-xy\right)\end{matrix}\right.\)
Đây là các bài hệ trong đề thi chuyên toán mong mọi người giúp vì mình bận quá nên không thể làm hết được ạ
1,ĐK: \(x,y\ne-2\)
HPT<=> \(\left\{{}\begin{matrix}x\left(x+2\right)+y\left(y+2\right)=\left(x+2\right)\left(y+2\right)\left(1\right)\\x^2\left(x+2\right)^2+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x^2\left(x+2\right)^2+2xy\left(x+2\right)\left(y+2\right)+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\\x^2\left(x+2\right)^2+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\end{matrix}\right.\)
=> \(2xy\left(x+2\right)\left(y+2\right)=0\)
<=>\(2xy=0\) (do x+2 và y+2 \(\ne0\))
<=> \(\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Tại x=0 thay vào (1) có: \(y\left(y+2\right)=2\left(y+2\right)\) <=> y= \(\pm2\) => y=2 (vì y khác -2)
Tại y=0 thay vào (1) có: \(x\left(x+2\right)=2\left(x+2\right)\) => x=2
Vậy HPT có 2 nghiệm duy nhất (2,0),(0,2)
2, ĐK: \(y\ne-1\)
HPT <=> \(\left\{{}\begin{matrix}x^2=2\left(x+3\right)\left(y+1\right)\left(1\right)\\\frac{3x^2}{y+1}=4-x\end{matrix}\right.\)
=> \(\frac{6\left(3+x\right)\left(y+1\right)}{y+1}=4-x\)
<=> 6(x+3)=4-x
<=> \(14=-7x\)
<=> \(x=-2\) thay vào (1) có \(4=2\left(y+1\right)\)
<=>y=1\(\)( tm)
Vậy hpt có một nghiệm duy nhất (-2,1)
3,\(\left\{{}\begin{matrix}x^2-y=y^2-x\left(1\right)\\x^2-x=y+3\left(2\right)\end{matrix}\right.\)
PT (1) <=> \(\left(x-y\right)\left(x+y\right)+\left(x-y\right)=0\)
<=> (x-y)(x+y+1)=0
<=>\(\left[{}\begin{matrix}x=y\\y=-x-1\end{matrix}\right.\)
Tại x=y thay vào (2) có \(y^2-y=y+3\) <=> \(y^2-2y-3=0\) <=> (y-3)(y+1)=0 <=> \(\left[{}\begin{matrix}y=3\\y=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Tại y=-1-x thay vào (2) có: \(x^2-x=-1-x+3\) <=> \(x^2=2\) <=> \(\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}y=-1-\sqrt{2}\\y=-1+\sqrt{2}\end{matrix}\right.\)
Vậy hpt có 4 nghiệm (3,3),(-1,-1), ( \(\sqrt{2},-1-\sqrt{2}\)),( \(-\sqrt{2},-1+\sqrt{2}\))
4,\(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=\frac{9}{2}\left(1\right)\\xy+\frac{1}{xy}+\frac{x}{y}+\frac{y}{x}=5\left(2\right)\end{matrix}\right.\)(đk:\(x\ne0,y\ne0\))
<=> \(\left\{{}\begin{matrix}\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)=\frac{9}{2}\\\left(y+\frac{1}{y}\right)\left(x+\frac{1}{x}\right)=5\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\frac{1}{x}=u\\y+\frac{1}{y}=v\end{matrix}\right.\)
Có \(\left\{{}\begin{matrix}u+v=\frac{9}{2}\\uv=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\v\left(\frac{9}{2}-v\right)=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\\left(v-\frac{5}{2}\right)\left(v-2\right)=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\\left[{}\begin{matrix}v=\frac{5}{2}\\v=2\end{matrix}\right.\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}v=\frac{5}{2}\\u=2\end{matrix}\right.\\\left[{}\begin{matrix}v=2\\u=\frac{5}{2}\end{matrix}\right.\end{matrix}\right.\)
Tại \(\left\{{}\begin{matrix}v=\frac{5}{2}\\u=2\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x+\frac{1}{x}=2\\y+\frac{1}{y}=\frac{5}{2}\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-2\right)\left(y-\frac{1}{2}\right)=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=1\\\left[{}\begin{matrix}y=2\\y=\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left[{}\begin{matrix}x=1\\y=\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
Tại \(\left\{{}\begin{matrix}v=2\\u=\frac{5}{2}\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x+\frac{1}{x}=\frac{5}{2}\\y+\frac{1}{y}=2\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}\left(x-2\right)\left(x-\frac{1}{2}\right)=0\\\left(y-1\right)^2=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\end{matrix}\right.\\y=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left[{}\begin{matrix}x=\frac{1}{2}\\y=1\end{matrix}\right.\end{matrix}\right.\)
Vậy hpt có 4 nghiệm (1,2),( \(1,\frac{1}{2}\)) ,( 2,1),(\(\frac{1}{2},1\)).
10.
\(\left\{{}\begin{matrix}2x^2-3xy+y^2+x-y=0\\x^2+x+1=y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-2xy-xy+y^2+x-y=0\\x^2+x+1=y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(2x-y+1\right)=0\\x^2+x+1=y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\y=2x+1\end{matrix}\right.\\x^2+x+1=y^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2+x+1=y^2\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\x^2+x+1=y^2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2+x+1=x^2\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\x^2+x+1=\left(2x+1\right)^2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\3x\left(x+1\right)=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=1\\\left[{}\begin{matrix}\left\{{}\begin{matrix}y=2x+1\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\x=-1\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y=-1\\\left\{{}\begin{matrix}x=0\\y=-\frac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=-1\\\left\{{}\begin{matrix}x=0\\y=-\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
@Võ Hồng Phúc vô giải
13.
\(\left\{{}\begin{matrix}x^2+y^2=5\\x^3+2y^3+\left(x-5\right)^2+\left(y+5\right)^2=55\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\x^3+2y^3+x^2+y^2-10x+10y+50=55\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\x^3+2y^3=10x-10y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\x^3+2y^3=2.5\left(x-y\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\x^3+2y^3=2.\left(x^2+y^2\right)\left(x-y\right)\left(1\right)\end{matrix}\right.\)
Từ \(\left(1\right)\) ta có: \(x^3-2x^2y+2xy^2-4y^3=0\)
\(\Leftrightarrow\left(x-2y\right)\left(x^2+2y^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2y=0\\x^2+2y^2=0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x^2+y^2=5\\x-2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y^2=5\\x=2y\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\\\left\{{}\begin{matrix}y=-1\\x=-2\end{matrix}\right.\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x^2+y^2=5\\x^2+2y^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\y^2=-5\end{matrix}\right.\text{ vô nghiệm}\)
Vậy hệ pt đã cho có 2 nghiệm \(\left(x;y\right)\in\left\{\left(1;2\right);\left(-1;-2\right)\right\}\)
12.
\(hpt\Leftrightarrow\left\{{}\begin{matrix}x^2+7=4y^2+4y\\\left(x+y\right)\left(x+2y+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+7=4y^2+4y\\\left[{}\begin{matrix}x=-y\\x=-2x-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+7=4y^2+4y\\x=-y\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+7=4y^2+4y\\x=-2x-1\end{matrix}\right.\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x^2+7=4y^2+4y\\x=-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y^2+7=4y^2+4y\\x=-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(y-3\right)\left(y+7\right)=0\\x=-y\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=3\\x=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y=-7\\x=7\end{matrix}\right.\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x^2+7=4y^2+4y\\x=-2x-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4y^2+4y+1+7=4y^2+4y\\x=-2x-1\end{matrix}\right.\text{ vô nghiệm}\)
Vậy hệ pt đã cho có 2 nghiệm ...
\(pt\Leftrightarrow\left\{{}\begin{matrix}x^2-5xy+7xy+6y^2+42-5y^2=42\\7xy+6y^2+42=x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=0\\7xy+6y^2+42=x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\-7y^2+6y^2+42=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y^2+y-42=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-y\\\left(y-6\right)\left(y+7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=6\\x=-6\end{matrix}\right.\\\left\{{}\begin{matrix}y=-7\\x=7\end{matrix}\right.\end{matrix}\right.\)
Vậy hệ pt có hai nghiệm ...
18.
\(\left\{{}\begin{matrix}x^2=\left(2-y\right)\left(2+y\right)\left(1\right)\\2x^3=\left(x+y\right)\left(4-xy\right)\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow2x^3=4x+4y-x^2y-xy^2\)
\(\Leftrightarrow2x^3=x\left(4-y^2\right)+y\left(4-x^2\right)\left(3\right)\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}x^2=4-y^2\\y^2=4-x^2\end{matrix}\right.\) Thay vào \(\left(3\right)\), ta được:
\(2x^3=x^3+y^3\)
\(\Leftrightarrow x^3=y^3\Leftrightarrow x=y\)
Thay vào \(\left(1\right)\), ta được:
\(x^2=4-x^2\Leftrightarrow2x^2=4\Leftrightarrow\left[{}\begin{matrix}x=y=\sqrt{2}\\x=y=-\sqrt{2}\end{matrix}\right.\)
Vậy hệ pt đã cho có 2 nghiệm ...
17.
Vì \(x=\pm y\) không phải là nghiệm của hệ pt nên
\(hpt\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=13\left(x-y\right)\\x^6-y^6=7\left(x^2-y^2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=13\left(x-y\right)\\\left(x^3-y^3\right)\left(x^3+y^3\right)=91\left(x^2-y^2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=13\left(x-y\right)\\13\left(x-y\right)\left(x^3+y^3\right)=91\left(x+y\right)\left(x-y\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(x^2+xy+y^2\right)=13\left(x-y\right)\\13\left(x-y\right)\left(x+y\right)\left(x^2-xy+y^2\right)=91\left(x+y\right)\left(x-y\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=13\\x^2-xy+y^2=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy=3\\x^2+y^2=10\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=10\\\left(x-y\right)^2+2xy=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=16\\\left(x-y\right)^2=4\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=4\\x-y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=-4\\x-y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=4\\x-y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=-4\\x-y=2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\end{matrix}\right.\)
Vậy hệ pt đã cho có 4 nghiệm ...
