计算两道整式的乘除(a-b+c-d)²-(a+b+c+d)²,用平方差计算(2x-y+1)²+(2x-y-1)²-2(2x-y+1)(2x-y-1),用完全平方计算只写一道也行呐
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![计算两道整式的乘除(a-b+c-d)²-(a+b+c+d)²,用平方差计算(2x-y+1)²+(2x-y-1)²-2(2x-y+1)(2x-y-1),用完全平方计算只写一道也行呐](/uploads/image/z/7575489-9-9.jpg?t=%E8%AE%A1%E7%AE%97%E4%B8%A4%E9%81%93%E6%95%B4%E5%BC%8F%E7%9A%84%E4%B9%98%E9%99%A4%28a-b%2Bc-d%29%26%23178%3B-%28a%2Bb%2Bc%2Bd%29%26%23178%3B%2C%E7%94%A8%E5%B9%B3%E6%96%B9%E5%B7%AE%E8%AE%A1%E7%AE%97%282x-y%2B1%29%26%23178%3B%2B%282x-y-1%29%26%23178%3B-2%282x-y%2B1%29%282x-y-1%29%2C%E7%94%A8%E5%AE%8C%E5%85%A8%E5%B9%B3%E6%96%B9%E8%AE%A1%E7%AE%97%E5%8F%AA%E5%86%99%E4%B8%80%E9%81%93%E4%B9%9F%E8%A1%8C%E5%91%90)
计算两道整式的乘除(a-b+c-d)²-(a+b+c+d)²,用平方差计算(2x-y+1)²+(2x-y-1)²-2(2x-y+1)(2x-y-1),用完全平方计算只写一道也行呐
计算两道整式的乘除
(a-b+c-d)²-(a+b+c+d)²,用平方差计算
(2x-y+1)²+(2x-y-1)²-2(2x-y+1)(2x-y-1),用完全平方计算
只写一道也行呐
计算两道整式的乘除(a-b+c-d)²-(a+b+c+d)²,用平方差计算(2x-y+1)²+(2x-y-1)²-2(2x-y+1)(2x-y-1),用完全平方计算只写一道也行呐
(a-b+c-d)²-(a+b+c+d)²
=(a-b+c-d+a+b+c+d)(a-b+c-d-a-b-c-d)
=(2a+2c)(-2b-2d)
=-4(a+c)(b+d)
(2x-y+1)²+(2x-y-1)²-2(2x-y+1)(2x-y-1),
=(2x-y+1-2x+y+1)²
=2²
=4
-4(a+b)(c+d)
-4(a+c)×(b+d)
=(A-B+C-D+A+B+C+D)(A-B+C-D-A-B-C-D)=(2A+2C)(-2B-2D)=-4(A+C)(B+D)