Question

[Trần Thị Tú Anh]

tìm x, y biết x⁴+y⁴=17 và x+y=3

Answer 1

[Trà My]

\(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=\left[\left(x+y\right)^2-2xy\right]^2-2x^2y^2\)

\(=\left(9-2xy\right)^2-2x^2y^2=81-36xy+4x^2y^2-2x^2y^2=81-36xy+2x^2y^2=17\)

<=>\(81-36xy+2x^2y^2-17=0\)<=>\(64-36xy+2x^2y^2=0\)

<=>\(2\left(x^2y^2-18xy+32\right)=0\)<=>\(2\left[\left(xy-9\right)^2-49\right]=0\)

<=>\(\left(xy-9\right)^2-49=0\Leftrightarrow\left(xy-9\right)^2=49\)

<=>\(\orbr{\begin{cases}xy-9=-7\\xy-9=7\end{cases}\Leftrightarrow}\orbr{\begin{cases}xy=2\\xy=16\end{cases}}\)

+) Với xy=2

Có: \(x+y=3\Leftrightarrow x=3-y\Leftrightarrow xy=3y-y^2=2\Leftrightarrow3y-y^2-2=0\)

\(\Leftrightarrow y^2-3y+2=0\Leftrightarrow\left(y-2\right)\left(y-1\right)=0\Leftrightarrow\orbr{\begin{cases}y=2\\y=1\end{cases}}\)

<=> Với y=2 thì x=1 hoặc y=1 thì x=2

+) Với xy=16

\(xy=3y-y^2=16\Leftrightarrow3y-y^2-16=0\Leftrightarrow y^2-3y+16=0\)

<=>\(\left(y-\frac{3}{2}\right)^2+\frac{55}{4}=0\Leftrightarrow\left(y-\frac{3}{2}\right)^2=-\frac{55}{4}\)

pt vô nghiệm vì \(\left(y-\frac{3}{2}\right)^2\ge0\)

Vậy ...............................