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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