x+x:0.2+x:0.04=62
0,03< x < 0.04 X = ....
0,031 ; 0,032 ; 0,033 ; 0,034 ; 0,035 ; ... ; 0,039
1000 x 0.04 - 100 : 2.5 + 10 x 0.4
1000 x 0.04 - 100 : 2.5 + 10 x 0.4
= 40 - 40 + 4
= 0 + 4
= 4
Bài làm :
1 000 x 0,04 - 100 : 2,5 + 10 x 0,4
=40 - 40 + 4
=0 + 4
=4
với x=? ta có 5x=0.04
y : 0.04 + y x 3/2 - y x 6.5 = 2014
tìm y
y:0,04 + y x 3/2-y x 6,5 = 2014
y x 25 + y x 1,5 -y x 6,5=2014
y(25+1,5-6,5) =2014
20 x y=2014
Vậy y = 2014/20= 100,7
tìm x
(2/11*13+2/13*15 + ... 2/19*21)*462 - [0.04/(x+1.05)] /0.12=19
`(2/(11.13)+2/(13.15)+....+2/(19.21)).462-[0,04/(x+1,05):0,12=19`
`=>(1/11-1/13+1/13-1/15+....+1/19-1/21).462-(2/(50(x+1,05)).25/3=19`
`=>(1/11-1/21).462-1/(3(x+1,05))=19`
`=>10/231. 462-1/(3x+3,15)=19`
`=>20-1/(3x+3,15)=19
`=>1/(3x+3,15)=1`
`=>3x+3,15=1`
`=>3x=-2,15`
`=>x=-43/60`
Vậy `x=-43/60`
Giải:
\(\left(\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{19.21}\right).462-\left[\dfrac{0,04}{x+1,05}:0,12\right]=19\)
\(\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{19}-\dfrac{1}{21}\right).462-\left[\dfrac{1}{\dfrac{25}{x+\dfrac{21}{20}}}:\dfrac{3}{25}\right]=19\)
\(\left(\dfrac{1}{11}-\dfrac{1}{21}\right).462-\left[\dfrac{1}{25.\left(x+\dfrac{21}{20}\right)}:\dfrac{3}{25}\right]=19\)
\(20-\left[\dfrac{25.1}{3.25.\left(x+\dfrac{21}{20}\right)}\right]=19\)
\(\dfrac{1}{3.\left(x+\dfrac{21}{20}\right)}=20-19\)
\(\dfrac{1}{3.x+\dfrac{63}{20}}=1\)
\(1:\left(3.x+\dfrac{63}{20}\right)=1\)
\(3.x+\dfrac{63}{20}=1:1\)
\(3.x+\dfrac{63}{20}=1\)
\(3.x=1-\dfrac{63}{20}\)
\(3.x=\dfrac{-43}{20}\)
\(x=\dfrac{-43}{20}:3\)
\(x=\dfrac{-43}{60}\)
Tính nhanh:
0.8 x 0.04 x 1.25 x 25 x 0.6524 x 0.3476
Phần
10 x 125 x 4x 25 x8
0,8 x 0,04 x 1,25 x 25 x 0,6524 x 0,3476
= ( 0,8 x 1,25 ) x ( 0,04 x 25 ) x ( 0,3476 + 0,5624 )
= 1 x 1 x 0,9
= 1 x 0,9
= 0,9
10 x 125 x 4 x 25 x 8
= 10 x ( 125 x 8 ) x ( 4 x 25 )
= 10 x 1 000 x 100
= 1 000 000
0,8.0,04.1,25.25.0,6524.0,3476
=(0,8.1,25).(0,04.25).(0,3476+0,5624)
=1.1.0,9
=0,9
10.125.4.25.8
=10.(125.8).(4.25)
=10.1 000.100
=1000000
#HỌC TỐT#
54 x 125 x (2,5)-5 x 0.04
viết dưới dạng lũy thừa
nung hh gom 0.12 mol Al , 0.04 Mol Fe3O4 thu duoc ran X . Cho X + dd HCl du thu duoc 0.15 Mol H2 va xg muoi . Tinh x
x gồm 0.1 mol Mg, 0.04 mol Al, 0.15 mol Zn, X+ Hno3 loãn dư, sau pư m tăng 13,23g. tinh số mol hno3
Ta có:mKL=13,23g=m tăng->ko có khí thoát ra,nên spk duy nhất là NH4NO3.Sau khi viêt quá trình cho,nhận e ta lại có:nNH4NO3=(nMg*2+nAL*3+nZn*2)/8=0,0775mol
Mà nHNO3=10nNH4NO3=0,775mol