Tính
7/8 + 0,5 + 3/4 + 0,125 - 75/100
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1/2:0,5-1/4:0,2+1/8:0,125+75%
=0,5:0,5-0,25:0,2+0,125+0,75
= 1 + 1,25 + 1 + 0,75
= 1 + 1 - 1,25 + 0,75
= 2 - 1,25 +0,75
= 0,75 + 0,75
= 1,5
1/2:0,5-1/4:0,2+1/8:0,125+75%=0,5:0,5-0,05+0,125:0,125+0,75=1-0,05+1+0,75=2,7
Không khó đâu nhé em:
\(75\%+\frac{1}{4}-\frac{1}{4}:0,25+\frac{1}{8}:0,125-\frac{1}{2}:0,5\)
\(=\frac{3}{4}+\frac{1}{4}-\frac{1}{4}:\frac{1}{4}+\frac{1}{8}:\frac{1}{8}-\frac{1}{2}:\frac{1}{2}\)
\(=\frac{3}{4}+\frac{1}{4}-1+1-1\)
\(=1-1+1-1\)
\(=0\)
\(\frac{1}{2}:0,5-\frac{1}{4}:0,25+\frac{1}{8}:0,125+75\%\)
\(=\frac{1}{2}\cdot2-\frac{1}{4}\cdot4+\frac{1}{8}\cdot8+0,75\)
\(=1+1+1+0,75\)
\(=3,75\)
2:
=1-1+1-1=0
3:
a: =>34*(100+1)/2:a=17
=>a=101
b: =>5/3(x-1/2)=5/4
=>x-1/2=5/4:5/3=3/4
=>x=5/4
1a, \(\dfrac{2005}{2001}\) = 1+\(\dfrac{4}{2001}\); \(\dfrac{2009}{2005}\)=1+\(\dfrac{4}{2005}\)vì\(\dfrac{4}{2001}\)>\(\dfrac{4}{2005}\)nên\(\dfrac{2005}{2001}\)>\(\dfrac{2009}{2005}\)
1b,\(\dfrac{1313}{1515}\)=\(\dfrac{1313:101}{1515:101}\)= \(\dfrac{13}{15}\); \(\dfrac{131313}{151515}\)=\(\dfrac{131313:10101}{151515:10101}\)=\(\dfrac{13}{15}\)
Vậy \(\dfrac{13}{15}\)=\(\dfrac{1313}{1515}\)=\(\dfrac{131313}{151515}\)
chỉ có một lần cân bằng cân Rôbecvan và 1 quả cân 1kg làm thế nào ta có thể lấy 2kg gạo ra khỏi bao gạo chứa 5kg
1/2:0,5-1/4:0,2+1/8:0,125+75%
= 50%:50%-25%:20%+12,5%:12.%5+75%
=1-25%:20%+12,5%+12.%5+75%
=75,75%
`a)1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10`
`=(1/10+9/10)+(2/10+8/10)+(3/10+7/10)+(4/10+6/10)+5/10`
`=10/10 + 10/10+10/10+10/10+1/2`
`=1+1+1+1+1/2`
`=2+2+1/2`
`=4+1/2`
`=8/2+1/2`
`=9/2`
__
`13,25:0,5+13,25:0,25+13,25:0,125+13,25×6`
`=13,25×1/(0,5)+13,25×1/(0,25)+13,25×1/(0,125)+13,25×6`
`=13,25×(1/(0,5)+1/(0,25)+1/(0,125)+6)`
`=13,25×(2+4+8+6)`
`=13,25×20`
`=265`
`#QiN`
\(\frac{7}{8}+0,5+\frac{3}{4}+0,125-\frac{75}{100}\)
\(=\frac{7}{8}+\frac{1}{2}+\frac{3}{4}+\frac{1}{8}-\frac{3}{4}\)
\(\left(\frac{7}{8}+\frac{1}{8}\right)+\left(\frac{3}{4}-\frac{3}{4}\right)+\frac{1}{2}\)
\(\frac{8}{8}+0+\frac{1}{2}\)
\(1+0+\frac{1}{2}\)
\(1+\frac{1}{2}\)
\(\frac{2}{2}+\frac{1}{2}\)
\(=\frac{3}{2}\)