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از اتحاد مزدوج در تجزیهٔ عبارت های جبری نیز استفاده می شود.
\({A^2} - {B^2} = (A + B)(A - B)\)
با توجه به این تساوی، جای خالی را پر کنید.
1 \({x^2} - 9 = (x + 3)(\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,)\)
2 \(4{y^2} - \frac{1}{4}{z^4} = (\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,)(\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,)\)
3 \({(2x + 1)^2} - {y^2} = \left[ {(2x + 1) - \,\,\,\,\,\,\,\,\,\,\,} \right]\left[ {(\,\,\,\,\,\,\,\,\,\,\,) + y} \right]\)
4
\(\begin{array}{l}1 - {(3a + z)^2} = \left[ {1 - (\,\,\,\,\,\,\,\,\,\,\,)} \right]\left[ {1 + (\,\,\,\,\,\,\,\,\,\,\,)} \right]\\\\ = (\,\,\,\,\,\,\,\,\,\,\,)(1 + 3a + z)\end{array}\)
5
\(\begin{array}{l}{(2x + 1)^2} - {(3x + 4)^2} = \\\\\left[ {(\,\,\,\,\,\,\,\,\,\,\,) - (\,\,\,\,\,\,\,\,\,\,\,)} \right]\left[ {(\,\,\,\,\,\,\,\,\,\,\,) + (\,\,\,\,\,\,\,\,\,\,\,)} \right]\\\\ = ( - x - 3)(\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,)\end{array}\)
6
\(\begin{array}{l}{x^4} - {y^4} = ({x^2} + {y^2})(\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,)\\\\ = ({x^2} + {y^2})(x + y)(\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,)\end{array}\)
1 \({x^2} - 9 = (x + 3)(x - 3)\)
2 \(4{y^2} - \frac{1}{4}{z^4} = (2y + \frac{1}{2}z)(2y - \frac{1}{2}z)\)
3 \({(2x + 1)^2} - {y^2} = \left[ {(2x + 1) - y} \right]\left[ {(2x + 1) + y} \right]\)
3 \({(2x + 1)^2} - {y^2} = \left[ {(2x + 1) - y} \right]\left[ {(2x + 1) + y} \right]\)
4
\(\begin{array}{l}1 - {(3a + z)^2} = \left[ {1 - (3a + z)} \right]\left[ {1 + (3a + z)} \right]\\\\ = (1 - 3a - z)(1 + 3a + z)\end{array}\)
5
\(\begin{array}{l}{(2x + 1)^2} - {(3x + 4)^2} = \\\\\left[ {(2x + 1) - (3x + 4)} \right]\left[ {(2x + 1) + (3x + 4)} \right]\\\\ = ( - x - 3)(5x + 5)\end{array}\)
6
\(\begin{array}{l}{x^4} - {y^4} = ({x^2} + {y^2})({x^2} - {y^2})\\\\ = ({x^2} + {y^2})(x + y)(x - y)\end{array}\)
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