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`(13+x)/20 = 3/4`
`(13+x) xx4=3xx20`
`(13+x)xx4=60`
`13+x=60:4`
`13+x=15`
`x=15-13`
`x=2`
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`(23-x)/25 =4/5`
`(23-x)xx5=4xx25`
`(23-x)xx5=100`
`23-x=100:5`
`23-x=20`
`x=23-20`
`x=3`
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do dịch covid mình quên dạng này rồi ://
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4\(\dfrac{1}{2}\)=4,5 3\(\dfrac{4}{5}\)=3,8 2\(\dfrac{3}{4}\)=2,75 1\(\dfrac{12}{25}\)=1,48.
4 1/2=4,5
3 4/5=3,8
2 3/4=2,75
1 12/25=1,48
đúng 100% nha
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`a)1/7xx2/7+1/7xx5/7+6/7`
`=1/7xx(2/7+5/7)+6/7`
`=1/7xx1+6/7`
`=1/7+6/7=1`
`b)6/11xx4/9+6/11xx7/9-6/11xx2/9`
`=6/11xx(4/9+7/9-2/9)`
`=6/11xx9/9`
`=6/11`
Sorry nãy ghi thiếu.
`c)4/25xx5/8xx25/4xx24`
`=(4xx5xx25xx24)/(25xx8xx4)`
`=(4xx5xx24)/(4xx8)`
`=(5xx24)/8`
`=5xx3=15`
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`a,3/10=0,3`
`3/100=0,03`
`4 25/100=4 1/4=4,25`
`2002/1000=2,002`
`b,1/4=0,25`
`3/5=0,6`
`7/8=0,875`
`1 1/2=1,5`
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{120}{121}.\dfrac{143}{144}\)
= \(\dfrac{1.3.2.4.3.5.4.6....10.12.11.13}{2^2.3^2.4^2.5^2...11^2.12^2}\)
= \(\dfrac{1.2.12.13}{2^2.12^2}=\dfrac{13}{2.12}=\dfrac{13}{24}\)
b) \(B=\dfrac{5}{9}.\dfrac{21}{25}.\dfrac{45}{49}.\dfrac{77}{81}....\dfrac{357}{361}.\dfrac{437}{441}\)
= \(\dfrac{1.5.3.7.5.9.7.11.....17.21.19.23}{3^2.5^2.7^2....19^2.21^2}=\dfrac{1.3.21.23}{3^2.21^2}\)
= \(\dfrac{23}{3.21}=\dfrac{23}{63}\)
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\(A\cdot\left(1-\dfrac{1}{4}\right)\cdot\left(1-\dfrac{1}{9}\right)\cdot\left(1-\dfrac{1}{16}\right)\left(1-\dfrac{1}{25}\right)=1\dfrac{3}{5}\)
=>\(A\cdot\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{5}\right)=\dfrac{8}{5}\)
=>\(A\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot\dfrac{6}{5}=\dfrac{8}{5}\)
=>\(A\cdot\dfrac{1}{5}\cdot\dfrac{6}{2}=\dfrac{8}{5}\)
=>\(A\cdot3=8\)
=>A=8/3
\(\dfrac{5}{25}+\dfrac{4}{5}=\dfrac{1}{5}+\dfrac{4}{5}=\dfrac{1+4}{5}=\dfrac{5}{5}=1\)
25/125+100/125=125/125=1