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Ta xét hệ \(\hept{\begin{cases}2x^2+xy-y^2-5x+y+2=0\left(1\right)\\x^2+y^2+x+y-4=0\left(2\right)\end{cases}}\)
Ta có: \(\left(1\right)\Leftrightarrow y^2-\left(x+1\right)y-2x^2+5x-2=0\)
\(\Leftrightarrow\left[y-\frac{x+1}{2}\right]^2-\left[\frac{\left(x+1\right)^2}{4}+2x^2-5x+2\right]=0\)
\(\Leftrightarrow\left[y-\frac{x+1}{2}\right]^2-\frac{9x^2-18x+9}{4}=0\)\(\Leftrightarrow\left[y-\frac{x+1}{2}\right]^2-\left(\frac{3x-3}{2}\right)^2=0\)
\(\Leftrightarrow\left(y-\frac{x+1}{2}-\frac{3x-3}{2}\right)\left(y-\frac{x+1}{2}+\frac{3x-3}{2}\right)=0\)\(\Leftrightarrow\left(y-2x+1\right)\left(y+x-2\right)=0\Leftrightarrow\orbr{\begin{cases}y-2x+1=0\\y+x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}y=2x-1\\y=2-x\end{cases}}\)
TH1: \(y=2x-1\), thay vào phương trình (2), ta được: \(x^2+\left(2x-1\right)^2+x+2x-1-4=0\)
\(\Leftrightarrow5x^2-x-4=0\Leftrightarrow\orbr{\begin{cases}x=1\Rightarrow y=1\\x=-\frac{4}{5}\Rightarrow y=\frac{-13}{5}\end{cases}}\)
TH2: \(y=2-x\), thay vào phương trình (2), ta được: \(x^2+\left(2-x\right)^2+x+2-x-4=0\)
\(\Leftrightarrow2x^2-4x+2=0\Leftrightarrow2\left(x-1\right)^2=0\Leftrightarrow x=1\Rightarrow y=1\)
Vậy hệ có 2 nghiệm \(\left(x;y\right)\in\left\{\left(1;1\right);\left(-\frac{4}{5};-\frac{13}{5}\right)\right\}\)
\(+,2x^2+xy-y^2-5x+y+2=0\)
\(\Leftrightarrow x^2+\frac{xy}{2}-\frac{y^2}{2}-\frac{5x}{2}+\frac{y}{2}+1=0\)
\(\Leftrightarrow x^2+x\left(\frac{y}{2}-\frac{5}{2}\right)-\frac{y^2}{2}+\frac{y}{2}+1=0\)
\(\Leftrightarrow x^2+2x.\frac{y-5}{4}+\left(\frac{y-5}{4}\right)^2-\left(\frac{y-5}{4}\right)^2-\frac{y^2}{2}+\frac{y}{2}+1=0\)
\(\Leftrightarrow\left(x+\frac{y-5}{4}\right)^2-\frac{y^2-10y+25}{16}-\frac{y^2}{2}+\frac{y}{2}+1=0\)
\(\Leftrightarrow\left(x+\frac{y-5}{4}\right)^2-\frac{9y^2-18y+9}{16}=0\)
\(\Leftrightarrow\left(x+\frac{y-5}{4}\right)^2-\left(\frac{3y-3}{4}\right)^2=0\)
\(\Leftrightarrow\left(x+\frac{y-5}{4}-\frac{3y-3}{4}\right)\left(x+\frac{y-5}{4}+\frac{3y-3}{4}\right)=0\)
\(\Leftrightarrow\left(x+\frac{-y-1}{2}\right)\left(x+y+2\right)=0\)
\(\orbr{\begin{cases}x=\frac{y+1}{2}\\x=-y-2\end{cases}}\)
vậy ....
\(\hept{\begin{cases}2x^2+xy-y^2-5x+y+2=0\\x^2+y^2+x+y-4=0\end{cases}}\)
\(\Leftrightarrow2x^2+xy-y^2-5x+y+2=x^2+y^2+x+y-4\)
\(\Leftrightarrow x^2+xy-y^2-5x+y+2=y^2+x+y-4\)
\(\Leftrightarrow x^2+xy-y^2-5x+y=y^2+x+y-4-2\)
\(\Leftrightarrow x^2+xy-y^2-5x+y=y^2+x+y-6\)
\(\Leftrightarrow x^2+xy-y^2+y=y^2+x+y-6+5x\)
\(\Leftrightarrow x^2+xy-y^2+y=y^2+6x+y-6\)
\(\Leftrightarrow x^2+xy-y^2=y^2+6x-6\)
\(\Leftrightarrow x^2+xy=y^2+6x-6+y^2\)
\(\Leftrightarrow x^2+xy=2y^2+6x-6\)
\(\Leftrightarrow x\left(x+y\right)=2\left(y^2+3x-3\right)\)
\(\hept{\begin{cases}2x^2+xy-y^2-5x+y+2=0\\x^2+y^2+x+y-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}y^2-\left(x+1\right)y-2x^2+5x-2=0\\x^2+y^2+x+y-4=0\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(y+x-2\right)\left(y-2x+1\right)=0\\x^2+y^2+x+y-4=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y+x-2=0\\x^2+y^2+x+y-4=0\end{cases}}\)hoặc \(\hept{\begin{cases}y-2x+1=0\\x^2+y^2+x+y-4=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}\)hoặc \(\hept{\begin{cases}x=\frac{-4}{5}\\y=\frac{-13}{5}\end{cases}}\)và \(\hept{\begin{cases}x=1\\y=1\end{cases}}\)
Vậy hpt có 2 nghiệm (x;y)=\(\left(1;1\right);\left(\frac{-4}{5};\frac{-13}{5}\right)\)
`Answer:`
a) \(\left(\sqrt{2}+1\right)x-\sqrt{2}=2\)
\(\Leftrightarrow\left(\sqrt{2}+1\right)x=2+\sqrt{2}\)
\(\Leftrightarrow x=\frac{2+\sqrt{2}}{\sqrt{2}+1}\)
\(\Leftrightarrow x=\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
\(\Leftrightarrow x=\sqrt{2}\)
b) \(x^4+x^2-6=0\)
\(\Leftrightarrow x^4+3x^2-2x^2-6=0\)
\(\Leftrightarrow x^2.\left(x^2+3\right)-2\left(x^2+3\right)=0\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-2=0\\x^2+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{2}\\x^2=-3\text{(Vô lý)}\end{cases}}}\)
\(a,\hept{\begin{cases}\left(x-y\right)^2=1\\2x^2+2y^2-2xy-y=0\end{cases}}\)
Xét từng TH với x-y=1 và x-y=-1
\(b,\hept{\begin{cases}\left(x-1\right)\left(y+2\right)=0\\xy-3x+2y=0\end{cases}}\)
Xét từng TH x=1 và y=-2
Đặt \(\frac{1}{2x-y}\)= a, \(\frac{1}{x +y}\)= b, ta có \(\hept{\begin{cases}3a-6b=1\\a-b=0\end{cases}}\)
Giải hệ phương trình được a=\(\frac{-1}{3}\), b=\(\frac{-1}{3}\)