Question

\(\left\{{}\begin{matrix}x+y=1\\x^4-y^4=7x-y\end{matrix}\right.\)

Giai hệ pt

Answer 1

\(\left\{{}\begin{matrix}x+y=1\left(1\right)\\x^4+y^4=7x-y\left(2\right)\end{matrix}\right.\)

Từ (1) , ta có: y=1-x.

Khi đó (2) \(\Leftrightarrow x^4-\left(1-x\right)^4=7x-1+x\)

\(\Leftrightarrow x^4-1+4x-6x^2+4x^3-x^4=8x-1\)

\(\Leftrightarrow4x^3-6x^2-4x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{2}\\x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=-1\\y=\dfrac{3}{2}\\y=1\end{matrix}\right.\)

Vậy hpt có tập nghiệm là \(\left(x;y\right)=\left(2;-1\right);\left(-\dfrac{1}{2};\dfrac{3}{2}\right);\left(0;1\right).\)