a, Ta có : \(\left\{{}\begin{matrix}x^2+y^2=1\\x^2-y^2-x+y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=1\\\left(x-y\right)\left(x+y\right)-\left(x-y\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=1\\\left(x-y\right)\left(x+y-1\right)=0\end{matrix}\right.\)
TH1 : \(x-y=0\Rightarrow x=y\)
- Thay vào PT ( I ) ta được : \(x^2+x^2=2x^2=1\)
\(\Rightarrow x=y=\dfrac{\sqrt{2}}{2}\)
TH2 : \(x+y-1=0\)
- Kết hợp PT ( I ) ta được hệ : \(\left\{{}\begin{matrix}x+y=1\\\left(x+y\right)^2-2xy=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\-2xy=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\xy=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\end{matrix}\right.\)
Vậy hệ phương trình có tập nghiệm là \(S=\left\{\left(\dfrac{\sqrt{2}}{2};\dfrac{\sqrt{2}}{2}\right);\left(1;0\right);\left(0;1\right)\right\}\)
b.
Đặt \(\sqrt{x^2+7}=t>0\)
\(\Rightarrow t^2-\left(x+4\right)t+4x=0\)
\(\Delta=\left(x+4\right)^2-16x=\left(x-4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{x+4+x-4}{2}=x\\t=\dfrac{x+4-x+4}{2}=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+7}=x\left(x\ge0\right)\\\sqrt{x^2+7}=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+7=x^2\left(vô-nghiệm\right)\\x^2+7=16\end{matrix}\right.\)
\(\Rightarrow x=\pm3\)