數學簡便運算初一 (2 3 4 5)X(

2021-08-13 02:38:39 字數 4555 閱讀 5678

1樓:匿名使用者

(2+3+4+5)x(1/2+1/3+1/4+1/5)=1+2/3+2/4+2/5+3/2+1+3/4+3/5+2+4/3+1+4/5+5/2+5/3+5/4+1

=6+(2/3+4/3+5/3)+(2/4+3/4+5/4)+(3/2+5/2)+(2/5+3/5+4/5)

=6+3+2/3+2+1/2+4+1+4/5=16+2/3+1/2+4/5

=16+20/30+15/30+24/30=16+59/30

=17+29/30

2樓:匿名使用者

(2+3+4+5)x(1/2+1/3+1/4+1/5)=12x(1/2+1/3+1/4+1/5)=12x1/2+12x1/3+12x1/4+12x1/5=6+4+3+12/5

=13+2+2/5

=15+2/5

=15又2/5

3樓:楊柳輕揚

(2+3+4+5)x(1/2+1/3+1/4+1/5)=[(2+5)+(3+4)]x[(1/2+1/5)+(1/3+1/4)]

=(7+7)x(7/10+7/12)

=2x7x7x(1/10+1/12)

=2x7x7x11/60

=539/30

4樓:匿名使用者

(2+3+4+5)x(1/2+1/3+1/4+1/5)=(12+2) x(1/2+1/3+1/4+1/5)=12 x(1/2+1/3+1/4+1/5)+ 2 x(1/2+1/3+1/4+1/5)

=6+4+3+12/5+1+2/3+1/2+2/5=14+14/5+7/6

=14+2+4/5+7/6

=16+59/30

=16+1+29/30

=17+29/30

小學數學 計算:(1+1/2+1/3+1/4)x(1/2+1/3+1/4+1/5)x(1+1/2+1/3+1/4+1/5)x(1/2+1/3+1/4)

5樓:匿名使用者

(1+1/2+1/3+1/4)x(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)x(1/2+1/3+1/4)

中間該是減號.

設1/2+1/3+1/4+1/5=m,1/2+1/3+1/4=n原式=(1+n)*(m)-(1+m)*n

=m+nm-n-nm

=m-n

=1/5

6樓:匿名使用者

中間是減號吧

(1+1/2+1/3+1/4)(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)(1/2+1/3+1/4)

=[1+(1/2+1/3+1/4)][(1/2+1/3+1/4)+1/5]-[(1+1/5) + (1/2+1/3+1/4)] * (1/2+1/3+1/4)

令x=1/2+1/3+1/4)

則原式=(1+x)(x+1/5) - ((1+1/5)+x)*x= x + 1/5 +x*x +x/5 -[x*x +x +x/5]= x+1/5+x*x+x/5 -x*x -x -x/5= 1/5

7樓:發生以後

(1+1/2+1/3+1/4)×(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)×(1/2+1/3+1/4)

=1 x(1/2+1/3+1/4+1/5)+ (1/2+1/3+1/4) x(1/2+1/3+1/4+1/5) -[1 x(1/2+1/3+1/4)+(1/2+1/3+1/4+1/5)x(1/2+1/3+1/4)]

=1 x(1/2+1/3+1/4+1/5)+ (1/2+1/3+1/4) x(1/2+1/3+1/4+1/5) -1 x(1/2+1/3+1/4)-(1/2+1/3+1/4+1/5)x(1/2+1/3+1/4)

=(1/2+1/3+1/4+1/5)-(1/2+1/3+1/4)=1/5

8樓:騙了他嗎

(1+1/2+1/3+1/4)(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)(1/2+1/3+1/4)

=[1+(1/2+1/3+1/4)][(1/2+1/3+1/4)+1/5]-[(1+1/5) + (1/2+1/3+1/4)] * (1/2+1/3+1/4)

令x=1/2+1/3+1/4)

則原式=(1+x)(x+1/5) - ((1+1/5)+x)*x= x + 1/5 +x*x +x/5 -[x*x +x +x/5]= x+1/5+x*x+x/5 -x*x -x -x/5= 1/5

數學高手啊!!這到題怎麼做:2*3*4*5*(1/2+1/3+1/4+1/5) 用簡便演算法啊?

9樓:匿名使用者

=20*5+6*9

=100+54

=154

10樓:匿名使用者

乘法分配律

或先通分後約分

簡便計算(1+1/2+1/3+1/4)x(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)x(1/2+1/3+1/4)

11樓:匿名使用者

令a=1/2+1/3+1/4,b=1/2+1/3+1/4+1/5(1+1/2+1/3+1/4)x(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)x(1/2+1/3+1/4)

=(1+a)*b-(1+b)*a

=b+ab-a-ab

=b-a

=1/5

12樓:匿名使用者

為了簡述方便

假設1/2+1/3+1/4=a

那麼原式=(1+a)(a+1/5)-(1+a+1/5)a=a+1/5+a*a+a/5-a-a*a-a/5=1/5

13樓:匿名使用者

(1+1/2+1/3+1/4)(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)(1/2+1/3+1/4)

=(1+1/2+1/3+1/4)(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4)(1/2+1/3+1/4)-1/5(1/2+1/3+1/4)

=(1+1/2+1/3+1/4)(1/2+1/3+1/4+1/5-1/2-1/3-1/4)-1/5(1/2+1/3+1/4)

=(1+1/2+1/3+1/4)(1/5)-1/5(1/2+1/3+1/4)

=1/5*(1+1/2+1/3+1/4-1/2-1/3-1/4)

=1/5

14樓:匿名使用者

(1+1/2+1/3+1/4)x(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)x(1/2+1/3+1/4)

=(1+1/2+1/3+1/4)x[(1/2+1/3+1/4)+(1/5)]-[(1+1/2+1/3+1/4)+(1/5)]x(1/2+1/3+1/4)

=(1+1/2+1/3+1/4)x(1/2+1/3+1/4)+(1+1/2+1/3+1/4)x(1/5)-(1+1/2+1/3+1/4)x(1/2+1/3+1/4)-(1/5)x(1/2+1/3+1/4)

=(1+1/2+1/3+1/4)x(1/5)-(1/5)x(1/2+1/3+1/4)

=1/5

簡便運算:(1+1/2+1/3+1/4)*(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)*(1/2+1/3+1/4)

15樓:風欲止而樹靜

(1+1/2+1/3+1/4)×(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)×(1/2+1/3+1/4)

=(1/2+1/3+1/4)×(1/2+1/3+1/4+1/5)+(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)×(1/2+1/3+1/4)

=(1/2+1/3+1/4+1/5-1-1/2-1/3-1/4-1/5))×(1/2+1/3+1/4)+(1/2+1/3+1/4+1/5)

=-1*(1/2+1/3+1/4)+1/2+1/3+1/4+1/5=1/5

16樓:沅江笑笑生

(1+1/2+1/3+1/4)*(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)*(1/2+1/3+1/4)

解 設(1+1/2+1/3+1/4+1/5)=t原式可變形為

(t-1/5)(t-1)-t(t-1/5-1)=t-t+1/5

=1/5

簡便計算(1+2/1-3/1+4/1)×(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)x(1/2+1/3+1/4)

17樓:新野旁觀者

(1+1/2+1/3+1/4)×(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)x(1/2+1/3+1/4)

=(1/2+1/3+1/4+1/5)+(1/2+1/3+1/4)×(1/2+1/3+1/4+1/5)-(1/2+1/3+1/4)-(1/2+1/3+1/4+1/5)×(1/2+1/3+1/4)

=(1/2+1/3+1/4+1/5)-(1/2+1/3+1/4)=1/5

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