17,56 * 0,24 =
0,254 * 0,27 =
giúp mình với !
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A = \(\dfrac{13}{50}\) + 9% + 41/100 + 0,24
A = 0,26 + 0,09 + 0,41 + 0,24
A = (0,26 + 0,24) + ( 0,09 + 0,41)
A = 0,5 + 0,5
A = 1


PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)
Ta có: \(n_{NaOH}=0,3\cdot0,1=0,03\left(mol\right)=n_{HCl\left(p/ứ\right)}\)
+) Trường hợp 1: NaOH còn dư
PTHH: \(Al+NaOH+H_2O\rightarrow NaAlO_2+\dfrac{3}{2}H_2\uparrow\)
Ta có: \(n_{NaOH\left(dư\right)}=n_{Al}=\dfrac{0,27}{27}=0,01\left(mol\right)\)
\(\Rightarrow n_{HCl}=n_{NaOH\left(p/ứ\right)}=0,02\left(mol\right)\) \(\Rightarrow a=C_{M_{HCl}}=\dfrac{0,02}{0,2}=0,1\left(M\right)\)
+) Trường hợp 2: HCl còn dư
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Ta có: \(n_{Al}=0,01\left(mol\right)\) \(\Rightarrow n_{HCl\left(dư\right)}=0,03\left(mol\right)\)
\(\Rightarrow n_{HCl}=0,06\left(mol\right)\) \(\Rightarrow a=C_{M_{HCl}}=\dfrac{0,06}{0,2}=0,3\left(M\right)\)

0, 24 x X = 3
X = 3 : 0, 24
X = 12, 5
Đúng 100%
Đúng 100%
Đúng 100%

= 2016 x 0,24 + 0,75 x 2016 + 2016 x 0,01
= 2016 x ( 0,24 + 0,75 + 0,01 )
= 2016 x 1
= 2016
=2016 x 0,24 + 2016 x 0,75 + 2016 x 0,01
=2016 x ( 0,24 + 0,75 + 0,01 )
=2016 x 1
=2016

`@` `\text {Ans}`
`\downarrow`
`a)`
`13/50 + 9% + 41/100 + 0,24`
`= 0,26 + 0,09 + 0,41 + 0,24`
`= (0,26 + 0,24) + (0,09 + 0,41)`
`= 0,5 + 0,5`
`= 1`
`b)`
`2018 \times 2020 - 1/2017 + 2018 \times 2019`
`= 2018 \times (2020 + 2019) - 1/2017`
`= 2018 \times 4039 - 1/2017`
`= 8150702`
`c)`
`1/2 + 1/6 + 1/12 + 1/20 +1/30 +1/42`
`=`\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}\)
`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{6}-\dfrac{1}{7}\)
`=`\(1-\dfrac{1}{7}\)
`= 6/7`
\(a,\dfrac{13}{50}+9\%+\dfrac{41}{100}+0,24\\ 0,26+0,09+0,41+0,24\\ =\left(0,26+0,24\right)+\left(0,09+0,41\right)\\ =0,5+0,5\\ =1\\ b,2018\times2020-\dfrac{1}{2017}+2018\times2019\\ =2018\times\left(2020+2019\right)-\dfrac{1}{2017}\\ =2018\times4039-\dfrac{1}{2017}\\ =3150702-\dfrac{1}{2017}\\ c,\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\\ =1-\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}.........+\dfrac{1}{6}-\dfrac{1}{7}\\ =1-\dfrac{1}{7}\\ =\dfrac{6}{7}\)

\(\frac{13}{50}+9\%+\frac{41}{100}+0,24\)
\(=\frac{13}{50}+\frac{9}{100}+\frac{41}{100}+\frac{24}{100}\)
\(=\frac{13}{50}+\left[\frac{9}{100}+\frac{41}{100}+\frac{24}{100}\right]\)
\(=\frac{13}{50}+\frac{74}{100}=\frac{13}{50}+\frac{37}{50}=1\)
đễ
=26%+9%+41%+24%
=(26%+24%)+(9%+41%)
=50%+50%
=100%

25,8x1,5=38,7
16,25*6,7=108,875
0,24*4,7=1,128
7,826*4,5=35,217