Question

giải hệ phương trình:

a) \(\left\{{}\begin{matrix}\left(x-y\right)^2\cdot y=2\\x^3-y^3=19\end{matrix}\right.\)

Answer 1

Lời giải:

Vì \((x-y)^2.y=2\neq 0\Rightarrow x-y\neq 0\)

HPT \(\Leftrightarrow \left\{\begin{matrix} 19(x-y)^2.y=38\\ 2(x^3-y^3)=38\end{matrix}\right.\)

\(\Rightarrow 19y(x-y)^2=2(x^3-y^3)\)

\(\Leftrightarrow 19y(x-y)^2-2(x-y)(x^2+xy+y^2)=0\)

\(\Leftrightarrow (x-y)[19y(x-y)-2(x^2+xy+y^2)]=0\)

\(\Leftrightarrow 19y(x-y)-2(x^2+xy+y^2)=0\) (do $x-y\neq 0$)

\(\Leftrightarrow -2x^2-21y^2+17xy=0\)

\(\Leftrightarrow 2x^2-17xy+21y^2=0\)

\(\Leftrightarrow (x-7y)(2x-3y)=0\)

Nếu \(x=7y\Rightarrow 19=x^3-y^3=(7y)^3-y^3=342y^3\)

\(\Rightarrow y^3=\frac{1}{18}\Rightarrow y=\sqrt[3]{\frac{1}{18}}\Rightarrow x=\frac{7}{\sqrt[3]{18}}\)

Nếu \(x=\frac{3}{2}y\Rightarrow 19=x^3-y^3=(\frac{3}{2}y)^3-y^3=\frac{19}{8}y^3\)

\(\Rightarrow y^3=8\Rightarrow y=2\Rightarrow x=\frac{3}{2}y=3\)

Vậy \((x,y)=\left(\frac{1}{\sqrt[3]{18}}, \frac{7}{\sqrt[3]{18}}\right); (3,2)\)