Bài 1 : 

A=2+22+23+...+299+2100A=2+22+23+...+299+2100

⇒2A=22+23+24+...+2100+2101⇒2A=22+23+24+...+2100+2101

⇒A=2101−2⇒A=2101−2

B=3+32+33+...+399+3100B=3+32+33+...+399+3100

⇒3B=32+33+34+...+3100+3101⇒3B=32+33+34+...+3100+3101

Bài 2 :

2.Chứng minh rằng

2¹²+3¹²+2¹³+2¹⁴+3¹⁵ chia hết cho 7

⇒2B=3101−3⇒2B=3101−3

⇒B=3101−32

đề :

= 1/100 - (1 / 100.99 +1/99.98 + ...+ 1/3.2 +1/2.1 ) 

=1/100 - (1 /1.2 +1/ 2.3 +...+ 1/ 98.99 +1 / 99.100)

=1/100 -( 1- 1/ 2 +1/2 -1/3 +...+1/98 -1/99 +1/99 -1/100)

=1/100 - ( 1- 1/100)

=1/100 - 99 /100

= -98/100

= -49 /50

1/100.99 - 1/99.98 - 1/98.97 -...- 1/3.2 - 1/2.1
=-(1/100.99 + 1/99.98 + 1/98.97 +...+ 1/3.2 + 1/2.1)
=-(1/2.1+1/3.2 +...+1/98.97+ 1/99.98 +1/100.99 )
=-(1/1.2+1/2.3+1/3.4+...+1/97.98+ 1/98.99 +1/99.100)
=-(1/1-1/2+1/2-1/3+1/3......-1/98+1/98-1/99+1/99-1/100)
=-(1/1-1/100)=-99/100
 

C = 1/100 - 1/100.99 - 1/99.98 - 1/98.97 - ... - 1/3.2 - 1/2.1

C = 1/100 - (1/1.2 + 1/2.3 + ... + 1/98.99 + 1/99.100)

C = 1/100 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/98 - 1/99 + 1/99 - 1/100)

C = 1/100 - (1 - 1/100)

C = 1/100 - 99/100

C = -98/100 = -49/50

\(\frac{1}{100}-\frac{1}{100\cdot99}-\frac{1}{99\cdot98}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)

\(=\frac{1}{100}-\left(\frac{1}{100\cdot99}+\frac{1}{99\cdot98}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)

\(=\frac{1}{100}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\right)\)

\(=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(=\frac{1}{100}-\frac{99}{100}\)

\(=\frac{-49}{50}\)

a) \(26+173+74+27\)

\(=\left(26+74\right)+\left(173+27\right)\)

\(=100+200\)

\(=300\)

b) \(75\cdot37+89\cdot46+75\cdot52-89\cdot21\)

\(=75\cdot\left(37+52\right)+89\cdot\left(46-21\right)\)

\(=75\cdot89+89\cdot25\)

\(=89\cdot\left(75+25\right)\)

\(=89\cdot100\)

\(=8900\)

c) \(2^7:2^2+5^4:5^3\cdot2^4-3\cdot2^5\)

\(=2^{7-2}+5^{4-3}\cdot2^4-3\cdot2^5\)

\(=2^5+5\cdot2^4-3\cdot2^5\)

\(=2^4\cdot\left(2+5-3\cdot2\right)\)

\(=2^4\cdot\left(7-6\right)\)

\(=2^4\)

\(=16\)

d) \(100:\left\{250:\left[450-\left(4\cdot5^3-2^2\cdot25\right)\right]\right\}\)

\(=100:\left\{250:\left[450-\left(4\cdot5^3-4\cdot5^2\right)\right]\right\}\)

\(=100:\left[250:\left(450-4\cdot5^2\cdot4\right)\right]\)

\(=100:\left[250:\left(450-400\right)\right]\)

\(=100:\left(250:50\right)\)

\(=100:5\)

\(=20\)

a) \(\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{99}-\left(\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

đặt \(A=\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)

\(A=1-\frac{1}{99}\)

\(A=\frac{98}{99}\)

thay A vào, ta được :

\(\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)

b) \(\frac{2}{100.99}-\frac{2}{99.98}-...-\frac{2}{3.2}-\frac{2}{2.1}\)

\(=\frac{2}{100.99}-\left(\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\right)\)

đặt \(A=\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\)

\(A=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{98.99}\)

\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)

\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)

\(A=2.\left(1-\frac{1}{99}\right)\)

\(A=2.\frac{98}{99}\)

\(A=\frac{196}{99}\)

Thay A vào, ta được :

\(\frac{2}{100.99}-\frac{196}{99}=\frac{-19598}{9900}\)

\(10+2\cdot x=4^5:4^3\)

\(\Rightarrow10+2\cdot x=4^{5-3}\)

\(\Rightarrow10+2\cdot x=4^2\)

\(\Rightarrow10+2\cdot x=16\)

\(\Rightarrow2\cdot x=16-10\)

\(\Rightarrow2\cdot x=6\)

\(\Rightarrow x=\dfrac{6}{2}=3\)

__________________

\(2\cdot x-6^2:18=3\cdot2^2\)

\(\Rightarrow2\cdot x-36:18=12\)

\(\Rightarrow2\cdot x-2=12\)

\(\Rightarrow2\cdot x=12+2\)

\(\Rightarrow2\cdot x=14\)

\(\Rightarrow x=\dfrac{14}{2}=7\)

_________________

\(70-5\cdot\left(x-3\right)=3\cdot2\)

\(\Rightarrow70-5\cdot\left(x-3\right)=6\)

\(\Rightarrow70-5x+15=6\)

\(\Rightarrow-5x+15=6-70\)

\(\Rightarrow-5x+15=64\)

\(\Rightarrow-5x=64-15\)

\(\Rightarrow-5x=49\)

\(\Rightarrow x=-\dfrac{49}{5}\)