求过程及答案、应用平方差公式的!① 1998²-1997×1999 ②﹙a-2b+3c﹚﹙a+2b-3c﹚ ③﹙x+y+2﹚﹙x+y-2﹚ ④﹙-3a+2b﹚﹙3a+2b﹚ ⑤﹙2m+3n﹚﹙2m-3n﹚-﹙3m-2n﹚﹙3m+2n﹚ ⑥﹙-3a+m﹚﹙4b+n﹚=16b²-9a²,则m=?、 n=?
网友回答
①1998²-1997×1999 = 1998²-(1998-1) (1998+1) = 1998²-( 1998²-1)=1
②﹙a-2b+3c﹚﹙a+2b-3c﹚=[a-(2b-3c)][a+(2b-3c)=a²-(2b-3c)²=a²-4b²+12bc-9c²
③﹙x+y+2﹚﹙x+y-2﹚=(x+y)²-4=x²+2xy+y²-4
④﹙-3a+2b﹚﹙3a+2b﹚=(2b)²-(3a)²=4b²-9a²
⑤﹙2m+3n﹚﹙2m-3n﹚-﹙3m-2n﹚﹙3m+2n﹚=4m²-9n²-9m²+4n²=-4m²-4n²
⑥﹙-3a+m﹚﹙4b+n﹚=16b²-9a²,则m=4b 、 n=3a
======以下答案可供参考======
供参考答案1:
① 1998²-1997×1999
=1998²-(1998-1)×(1998+1)
=1998²-(1998²-1)
=1998²-1998²+1
=1②(a-2b+3c)(a+2b-3c)
=[a-(2b-3c)][a+(2b-3c)]
=a²-(2b-3c)²
=a²-(4b²-12bc+9c²)
=a²-4b²+12bc-9c²
③(x+y+2)(x+y-2)
=[(x+y)+2][(x+y)-2]
=(x+y)²-2²
=x²+2xy+y²-4
④(-3a+2b)(3a+2b)
=(2b)²-(3c)²
=4b²-9c²
⑤(2m+3n)(2m-3n)-(3m-2n)(3m+2n)
=(2m)²-(3n)²-[(3m)²-(2n)²]
=4m²-9n²-9m²+4n²
=-5m²-5n²
⑥∵(-3a+m)(4b+n)=16b²-9a²
=(4b)²-(3a)²
∴m=4b
n=3a