(2/3-4/7)×21+4/5简便计算
能简便的用简便方法计算.(2/3-4/7)*21+4/5=?(4/5+1/4)除以7/2-1/10=?28*[1/(3又1/10-3.09)]/0.4=?1-[1-(1-...
能简便的用简便方法计算.
(2/3-4/7)*21+4/5=?(4/5+1/4)除以7/2-1/10=?28*[1/(3又1/10-3.09)]/0.4=?
1-[1-(1-2/3除以2/3)]*3/4+5=? 展开
(2/3-4/7)*21+4/5=?(4/5+1/4)除以7/2-1/10=?28*[1/(3又1/10-3.09)]/0.4=?
1-[1-(1-2/3除以2/3)]*3/4+5=? 展开
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(2/3-4/7)*21+4/5=?
=2/3*21-4/7*21+4/5
=2*7-4*3+4/5
=14-12+4/5
=2+4/5
=2又4/5
=2.8
(4/5+1/4)除以7/2-1/10=?
=(4*4+1*5)/20 除以7/2-1/10 分子乘以20
=(16+5)/20除以7/2-1/10
=21/20除以7/2-1/10
=(21*2)/(20*7)-1/10
=3/10-1/10 分母21与分子7消得3;分母2与分子20消得1/10 =>3/10
=2/10
=0.2
28*[1/(3又1/10-3.09)]/0.4=?
=28*[1/(3.1-3.09)]/0.4 1/10=0.1 所以3又1/10=3.1 所以(3又1/10-3.09)=(3.1-3.09)
=28*[1/0.01]/0.4
=28*[1/(1/100)]/0.4 0.01=1/100
=28*100/0.4 1/(1/100)=100
=28/0.4*100 0.4=4/10; 28/(4/10)=28*10/4=28/4*10=7*10=70
=70*100
=7000
1-[1-(1-2/3除以2/3)]*3/4+5=?
=1-[1-(1-2/3*3/2)]*3/4+5 (2/3)/(2/3)=2/3*3/2=分子分母互消=1
=1-[1-(1-1)]*3/4+5
=1-[1-0]*3/4+5
=1-1*3/4+5
=1-3/4+5
=1/4+5
=5又1/4
=5.25
=2/3*21-4/7*21+4/5
=2*7-4*3+4/5
=14-12+4/5
=2+4/5
=2又4/5
=2.8
(4/5+1/4)除以7/2-1/10=?
=(4*4+1*5)/20 除以7/2-1/10 分子乘以20
=(16+5)/20除以7/2-1/10
=21/20除以7/2-1/10
=(21*2)/(20*7)-1/10
=3/10-1/10 分母21与分子7消得3;分母2与分子20消得1/10 =>3/10
=2/10
=0.2
28*[1/(3又1/10-3.09)]/0.4=?
=28*[1/(3.1-3.09)]/0.4 1/10=0.1 所以3又1/10=3.1 所以(3又1/10-3.09)=(3.1-3.09)
=28*[1/0.01]/0.4
=28*[1/(1/100)]/0.4 0.01=1/100
=28*100/0.4 1/(1/100)=100
=28/0.4*100 0.4=4/10; 28/(4/10)=28*10/4=28/4*10=7*10=70
=70*100
=7000
1-[1-(1-2/3除以2/3)]*3/4+5=?
=1-[1-(1-2/3*3/2)]*3/4+5 (2/3)/(2/3)=2/3*3/2=分子分母互消=1
=1-[1-(1-1)]*3/4+5
=1-[1-0]*3/4+5
=1-1*3/4+5
=1-3/4+5
=1/4+5
=5又1/4
=5.25
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