\(100+200=\)
\(10\times5=\)
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Đề sai chắc luôn đoạn kìa là `3xx5^{x-2}` mới đúng
`3xx5^{x-2}+4xx5^{x-3}=19xx5^10`
`=>3xx5^{x-3+1}+4xx5^{x-3}=19xx5^10`
`=>3xx5xx5^{x-3}+4xx5^{x-3}=19xx5^10`
`=>15xx5^{x-3}+4xx5^{x-3}=19xx5^10`
`=>19xx5^{x-3}=19xx5^10`
`=>5^{x-3}=5^10`
`=>x-3=10`
`=>x=13`
Vậy `x=13`
(7.x-11)3=22*52+200
(7.x-11)3=1000 = 103
suy ra: 7x-11=10
7x=10+11
7x=21
x=21:7=3
Ta có: B-A=1x3+2x4+3x5+4x6+...+100x102-(1x2+2x3+3x4+4x5+...+100x101)
=1x3+2x4+3x5+4x6+...100x102-1x2-2x3-3x4-4x5-...-100x101
=1+2+3+4+...+100
=((100-1):1+1)x((100-1):2)
=100x(101:2)
=5050
Có: A=\(\dfrac{3}{1.5}+\dfrac{3}{5.10}+...+\dfrac{3}{100.105}\)
=> A=\(3.\dfrac{5}{5}\left(\dfrac{1}{1.5}+\dfrac{1}{5.10}+...+\dfrac{1}{100.105}\right)\)
=> A= \(3.\dfrac{1}{5}\left(\dfrac{5}{1.5}+\dfrac{5}{5.10}+...+\dfrac{5}{100.105}\right)\)
=> A=\(\dfrac{3}{5}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{105}\right)\)
=> A= \(\dfrac{3}{5}\left(1-\dfrac{1}{105}\right)\)=\(\dfrac{3}{5}.\dfrac{104}{105}=\dfrac{312}{525}\)
1005 : 1003 = 1002
102 . 101 = 103
500200 . 50010 = 500210
105 . 102 = 107
10030 . 10010 = 10040
200100 : 20010 = 20090
10020 . 10010 = 10030
300100 . 300100 = 300200
1005 : 1003
= 1005 - 3
= 1002
102 . 101
= 102 + 1
= 103
500200 . 50010
= 500200 + 10
= 500210
105 . 102
= 105 + 2
= 107
10030 . 10010
= 10030 + 10
= 10040
200100 : 20010
= 200100 - 10
= 200190
10020 . 10010
= 10020 + 10
= 10030
300100 .300100
= 300100 + 100
= 300200
Ta có:
\(A=\frac{3}{1\cdot5}+\frac{3}{5\cdot10}+...+\frac{3}{100\cdot105}\)
\(=\frac{3}{5}\cdot\left(\frac{5}{1\cdot5}+\frac{5}{5\cdot10}+...+\frac{5}{100\cdot105}\right)\)
\(=\frac{3}{5}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{105}\right)\)
\(=\frac{3}{5}\left(1-\frac{1}{105}\right)=\frac{3}{5}\cdot\frac{104}{105}=\frac{312}{525}\)
300
50
\(100+200=300\)
\(10\times5=50\)