Question

решить систему уравнений

x+y=4
(x2-y2)(x-y)=144​

Answer 1

Решение:

$\begin {cases} x+y=4 \\ \left(x^2-y^2\right)(x-y)=144 \end.\\\\\left(x^2-y^2 \right)(x-y)=144 \\(x-y)(x+y)(x-y)=144\\(x+y)(x-y)^2=144\\\\\begin {cases} x+y=4 \\ 4(x-y)^2=144 \end. \\\\4(x-y)^2=144\\(x-y)^2=36\\x-y=\pm \sqrt{36}\\x-y=\pm 6\\\\\begin {cases} x+y=4 \\\begin {bmatrix} x-y=6 \\ x-y= -6 \end. \end. \Leftrightarrow$

$\Leftrightarrow\begin {cases} x=4-y \\\begin {bmatrix} 4-y-y=6 \\ 4-y-y= -6 \end. \end. \Leftrightarrow \quad\begin {cases} x=4-y \\\begin {bmatrix} 2y=-2 \\ 2y= 10 \end. \end. \Leftrightarrow \quad\begin {cases} x=4-y \\\begin {bmatrix} y=-1 \\ y= 5 \end. \end. \Leftrightarrow \\\\\Leftrightarrow \begin {bmatrix} \begin {cases} y=-1 \\ x=4+1 \end. \\ \begin {cases} y=5 \\ x=4-5 \end. \end. \Leftrightarrow \quad\begin {bmatrix} \begin {cases} y=-1 \\ x=5 \end. \\ \begin {cases} y=5 \\ x=-1 \end.\end.$

Ответ: ( 5; −1 ), ( −1; 5 ).