运用平方差公式计算: ①40又1/3×39又2/3 ②2014²-2015×293 ③9﹙10+1﹚﹙10²+1﹚+1运用平方差公式计算:①40又1/3×39又2/3②2014²-2015×293③9﹙10+1﹚﹙10²+1﹚+1④5﹙a-2b﹚﹙

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运用平方差公式计算: ①40又1/3×39又2/3 ②2014²-2015×293 ③9﹙10+1﹚﹙10²+1﹚+1运用平方差公式计算:①40又1/3×39又2/3②2014²-2015×293③9﹙10+1﹚﹙10²+1﹚+1④5﹙a-2b﹚﹙
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运用平方差公式计算: ①40又1/3×39又2/3 ②2014²-2015×293 ③9﹙10+1﹚﹙10²+1﹚+1运用平方差公式计算:①40又1/3×39又2/3②2014²-2015×293③9﹙10+1﹚﹙10²+1﹚+1④5﹙a-2b﹚﹙
运用平方差公式计算: ①40又1/3×39又2/3 ②2014²-2015×293 ③9﹙10+1﹚﹙10²+1﹚+1
运用平方差公式计算:
①40又1/3×39又2/3
②2014²-2015×293
③9﹙10+1﹚﹙10²+1﹚+1
④5﹙a-2b﹚﹙1/5a+2/5b﹚
⑤﹙﹣3a+b﹚﹙﹣2b+3a﹚+4b²
⑥﹙x+y﹚²-﹙x-y﹚²
⑦25²-24²
⑧﹙1-1/2²﹚﹙1-1/3²﹚﹙1-1/4²﹚...﹙1-1/9²﹚﹙1-1/10²﹚
⑨若﹙a+b﹚²=16,﹙a-b﹚²=8,求ab的值
⑩计算:﹙﹣a-2b﹚﹙2a-b﹚-﹙a-2b﹚﹙b+2a﹚
⑾﹙a+2b﹚﹙a-2b﹚+﹙a+2b﹚²-4ab
⑿﹙2x+y+1﹚﹙2x+y-1﹚
⒀﹙x-2y﹚﹙x+2y﹚-﹙x+2y﹚²
⒁﹙a+b+c﹚²
⒂﹙2x-y-1﹚²

运用平方差公式计算: ①40又1/3×39又2/3 ②2014²-2015×293 ③9﹙10+1﹚﹙10²+1﹚+1运用平方差公式计算:①40又1/3×39又2/3②2014²-2015×293③9﹙10+1﹚﹙10²+1﹚+1④5﹙a-2b﹚﹙
原式= (40+1/3)(39+2/3)=(40+1/3)(40-1/3)=1600-1/9
2014²-2015×293=2014²-2015×293-1+1
=(2014+1)(2014-1)-2015×293+1
=2015x2013 -2015×293+1
=2015(2013-293) + 1 = 2015x1720 + 1
3、解 9x11x101+1 = 99x101 + 1= (100-1)(100+1) +1 = 10000-1+1 = 10000
5﹙a-2b﹚﹙1/5a+2/5b﹚=5x1/5 (a - 2b)(a + 2b) =a² - 4b²
(﹣3a+b﹚﹙﹣2b+3a﹚+4b² = 4b² -(3a-b)(3a-2b)
= 4b² -(9a²-9ab+2b²)
=2b² -9a²+9ab
=(b+3a)(b-3a)+b(b+3a)+6ab
=(b+3a)(2b-3a)+6ab
﹙x+y﹚²-﹙x-y﹚² = (x+y+x-y)(x+y-x+y) = 4xy
25²-24² =(24+1)²-24² = (24+1+24)(24+1-24) = 49
﹙1-1/2²﹚﹙1-1/3²﹚﹙1-1/4²﹚...﹙1-1/9²﹚﹙1-1/10²﹚=(1+1/2)(1-1/2)(1+1/3)(1-1/3)(1+1/4)(1-1/4).(1+1/9)(1-1/9)(1+1/10)(1-1/10)
=(3/2)(1/2)(4/3)(2/3)(5/4)(3/4).(10/9)(8/9)(11/10)(9/10)
=[(3/2)(4/3)(5/4)..,(10/9)(11/10)][(1/2)(2/3)(3/4).(8/9)(9/10)]
=(11/2)(1/10) = 11/20
﹙a+b﹚²=16 ,a²+b²+2ab =16 (1)
﹙a-b﹚²=8 a²+b²-2ab =8 (2)
(1) - (2) 得:4ab = 8 解得:ab = 2
(﹣a-2b)(2a-b)-(a-2b)(b+2a) = (a+2b)(b-2a)-(a-2b)(b+2a)
=(-2a²-3ab+2b²) - (2a²-3ab-2b²) = 4b²-4a² = 4(b+a)(b-a)
太晚了,我在替你写作业啊!剩下的很简单,你仿照上面的方法可以做出来的!

1.40又1/3×39又2/3=(40+1/3)*(40-1/3)=40²-(1/3)²
4.原式=5*(a-2b)*(1/5)*(a+2b)=(a-2b)(a+2b)=a²-4b²