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- Với \(x=0\) không phải nghiệm
- Với \(x\ne0\):
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+\dfrac{y^2+1}{x}=2\\\left(x+y\right)^2-2\left(\dfrac{y^2+1}{x}\right)=-1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\\dfrac{y^2+1}{x}=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u+v=2\\u^2-2v=-1\end{matrix}\right.\)
\(\Rightarrow u^2-2\left(2-u\right)=-1\)
\(\Leftrightarrow u^2+2u-3=0\Rightarrow\left[{}\begin{matrix}u=1\Rightarrow v=1\\u=-3\Rightarrow v=5\end{matrix}\right.\)
\(\Rightarrow\) ... (bạn tự thế vào giải tiếp)
Nhận thấy \(x=y=0\) là 1 nghiệm
Với \(xy\ne0\) hệ tương đương:
\(\left\{{}\begin{matrix}\frac{1}{x^2}+\frac{1}{y^2}=2\\\left(\frac{x+y}{xy}\right)\left(\frac{1+xy}{xy}\right)=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(\frac{1}{x}+\frac{1}{y}\right)^2-\frac{2}{xy}=2\\\left(\frac{1}{x}+\frac{1}{y}\right)\left(1+\frac{1}{xy}\right)=4\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a=\frac{1}{x}+\frac{1}{y}\\b=\frac{1}{xy}\end{matrix}\right.\) với \(a^2\ge4b\)
\(\Rightarrow\left\{{}\begin{matrix}a^2-2b=2\\a\left(b+1\right)=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2-2\left(b+1\right)=0\\b+1=\frac{4}{a}\end{matrix}\right.\)
\(\Rightarrow a^2-\frac{8}{a}=0\Leftrightarrow a=3\Rightarrow b=\frac{1}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=3\\\frac{1}{xy}=\frac{1}{3}\end{matrix}\right.\) bạn tự giải nốt
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Ta có \(\left\{{}\begin{matrix}x^2+y^2=2x^2y^2\\\left(x+y\right)\left(1+xy\right)=4x^2y^2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left(x+y\right)^2=2\left(xy\right)^2+2xy\\\left(x+y\right)\left(1+xy\right)=4\left(xy\right)^2\end{matrix}\right.\)(1)
Đặt a=x+y,b=xy
Vậy (1)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a^2=2b^2+2b\left(3\right)\\a\left(1+b\right)=4b^2\left(2\right)\end{matrix}\right.\)
Trong phương trình (2), nếu 1+b=0\(\Leftrightarrow b=-1\)
Vậy \(\left(2\right)\Leftrightarrow a.0=4\left(ktm\right)\)
Vậy 1+b\(\ne0\)
Vậy (2)\(\Leftrightarrow a=\dfrac{4b^2}{1+b}\)
Thay vào (3)\(\Leftrightarrow\left(\dfrac{16b^2}{1+b}\right)=2b^2+2b\Leftrightarrow16b^4=\left(2b^2+2b\right)\left(b^2+2b+1\right)\Leftrightarrow16b^4=2b^4+4b^3+2b^2+2b^3+4b^2+2b\Leftrightarrow16b^4=2b^4+6b^3+6b^2+2b\Leftrightarrow14b^4-6b^3-6b^2-2b=0\Leftrightarrow7b^4-3b^3-3b^2-b=0\Leftrightarrow b\left(7b^3-3b^2-3b-1\right)=0\Leftrightarrow b\left(7b^3-7b^2+4b^2-4b+b-1\right)=0\Leftrightarrow b\left[7b^2\left(b-1\right)+4b\left(b-1\right)+\left(b-1\right)\right]=0\Leftrightarrow b\left(b-1\right)\left(7b^2+4b+1\right)=0\)(*)
Vì 7b2+4b+1>0
(*)\(\Leftrightarrow\)\(\left[{}\begin{matrix}b=0\\b=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}a=0\\a=2\end{matrix}\right.\)
TH1:a=0;b=0\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y=0\\xy=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
TH2:a=2;b=1\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy (x;y)={(0;0);(1;1)}
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 \(\left(x;y\right)=\left(1;2\right)\)
a.\(\left\{{}\begin{matrix}4x+2y=14\\2x-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=18\\2x-2y=4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=2\\4-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\-2y=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
vậy hệ pt có ndn \(\left\{2;0\right\}\)
b.\(\left\{{}\begin{matrix}2x-4y=0\\3x+2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=0\\6x+4y=16\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}8x=16\\2x-4y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\4-4y=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=2\\-4y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
vậy hệ pt có ndn \(\left\{2;1\right\}\)
