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Co Gai De Thuong
A = 2 + 22 + 23 + ... + 299 + 2100
= ( 2 + 22 + 23 + 24 + 25 ) + ... + ( 296 + 297 + 298 + 299 + 2100 )
= 2 x ( 1 + 2 + 22 + 23 + 24 ) + ... + 296 x ( 1 + 2 + 22 + 23 + 24 )
= 2 x 31 + ... + 296 x 31
= 31 ( 2 + ... + 296 )
Vậy A chia hết cho 31
A = 2 + 22 + 23 + 24 + 25 + .... + 296 + 297 + 298 + 299 + 2100
A = [2 + 22 + 23 + 24 + 25] + ... + 295[2 + 22 + 23 + 24 + 25]
A = 62 + ... + 295.62
A = 2.31 + .... + 295.2.31
A = 31.2.[20 + 25 + ... +295]
=> A \(⋮31\)
1, 15+(20-35)
=15+(-15)
=0
2,37.24+76.37
=37(24+76)
=37.100
=3700
3,420:{30:[260-(91.5-\(2^3\).\(5^2\))]}
=420:{30:[260-(455-200)
=420:[30:(260-455+200)
=420:(30:5)
=420:6
=70
5,2(x-11):3=56
<=>2(x-11)=56.3
<=>2(x-11)=168
<=>x-11=168:2
<=>x-11=84
<=>x=84+11
<=>x=95
Vậy x=95
a) 128 - 3(x + 4) = 23
3(x + 4) = 128 -23
3(x + 4) = 105
x + 4 = 105 : 3
x + 4 = 35
x = 35 - 4
x = 31
\(a.128-3\left(x+4\right)=23.\)
\(3\left(x+4\right)=128-23\)
\(3\left(x+4\right)=105\)
\(x+4=105:3=35\)
\(x=35-4=31\)
\(b.\left[\left(4x+28\right)\cdot3+55\right]:5=35\)
\(\left(4x+28\right)\cdot3+55=35\cdot5=175\)
\(\left(4x+28\right)\cdot3=175-55=120\)
\(4x+28=120:3=40\)
\(4x=40-28=12\)
\(x=12:4=3\)
2x x 16 = 128
2x = 128 : 16
2 x = 8
2x = 23
3x : 9 = 27
3x = 27 x 9
3x =243
3x = 35
[ 2x + 1 ]3 = 27
2x3 + 13 = 27
2x3 +1 = 27
2x3 = 27 - 1
2x3 = 26
\(a,\dfrac{7}{12}-\left(x+\dfrac{7}{10}\right):\dfrac{6}{5}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7}{12}-x-\dfrac{7}{10}:\dfrac{6}{5}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7}{12}-x-\dfrac{7}{12}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7}{12}-x=\dfrac{5}{4}+\dfrac{7}{12}\)
\(\Leftrightarrow\dfrac{7}{12}-x=\dfrac{11}{6}\)
\(\Leftrightarrow x=\dfrac{7}{12}-\dfrac{11}{6}\)
\(\Leftrightarrow\dfrac{-5}{4}\)
\(x^4\cdot x^7\cdot...\cdot x^{100}\)
\(=x^{4+7+...+100}\)
\(=x^{52\cdot33}=x^{1716}\)
\(x^1\cdot x^2\cdot x^3\cdot...\cdot x^{2006}\)
Ta có : \(x^1\cdot x^2=x^{1+2}=x^3\)
Tương tự : \(x^1\cdot x^2\cdot x^3=x^{1+2+3}=x^6\)
Áp dụng vào bài toán :
\(x^1\cdot x^2\cdot x^3\cdot...\cdot x^{2006}=x^{1+2+3+...+2006}\)
\(\Rightarrow x^{1+2+3+...+2006}=x^{2013021}\)
\(B=12.10^8+17.10^7+5.10^4+3\)
\(B=12.10^4.10^4+17.10^4.10^3+5.10^4+3\)
\(B=12.10000.10^4+17.1000.10^4+5.10^4+3\)
\(B=120000.10^4+17000.10^4+5.10^4+3\)
\(B=10^4\left(120000+17000+5\right)+3\)
\(B=10^4.137005+3\)
\(B=1370050000+3=1370050003\)
thank you so much