Ta có: A = 1.2 + 2.3 + 3.4 + 4.5 +.....+ 98.99

=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + ..... +98.99.(100 - 97)

=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 98.99.100

=> 3A = 98.99.100

=> A = 98.99.100 / 3

=> A = 323400

a) 9 + 99 + 999 + ... + 999999

= (10 - 1) + (100 - 1) + (1000 - 1) + ... + (1000000 - 1)

= (10¹ + 10² + 10³ + ... + 10⁶) - (1.6)

= 1111110 - 6 = 1111104

b) 1 + 11 + 111 + ... + 1111111

= 1 + (10¹ + 1) + (10² + 10¹ + 1) + ... + (10⁶ + 10⁵ + 10⁴ + 10³ + 10² + 10¹ + 1)

= 10¹ . 6 + 10² . 5 + 10³ . 4 + ... + 10⁶. 1) + (1 + 1.6)

= 60 + 500 + 4000 + ... + 1000000 + 7

= 1234560 + 7 = 1234567

c) C = 1.2 + 2.3 + 3.4 + 4.5 + ... + 98.99

3C = 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 + ... + 98.99.3

3C = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 98.99.(100 - 97)

3C = 1.2.3 + 2.3.4 - 2.3.1 + 3.4.5 - 3.4.2 +...+ 98.99.100 - 98.99.97

3C = 98.99.100

C = \(\dfrac{98.99.100}{3}\) = 323400

d) D = 1.3.5 + 3.5.7 + 5.7.9 + ... + 95.97.99

8D = 1.3.5.8 + 3.5.7.8 + 5.7.9.8 + ... + 95.97.99.8

8D = 1.3.5.(7 + 1) + 3.5.7.(9 - 1) + 5.7.9.(11 - 3) + ... + 95.97.99.(101 - 93)

8D = 1.3.5.7 + 1.3.5.1 + 3.5.7.9 - 3.5.7.1 + 5.7.9.11 - 5.7.9.3 + ... + 95.97.99.101 - 95.97.99.93

8D = 1.3.5.1 + 95.97.99.101

D = \(\dfrac{1.3.5.1+95.97.99.101}{8}=15517600\)

2.
a) (2x + 1)³ = 125
    (2x + 1)³ = 5³
     2x + 1    = 5
     2x          = 5 - 1
     2x          = 4
       x          = 4:2
       x          = 2
Vậy x = 2
b) 5^x+1 = 5⁴
    x + 1 = 4
    x       = 4 - 1 
    x       = 3
Vây x = 3

a) \(A=1.2+2.3+3.4+...+98.99+99.100\)

\(3A=1.2.3+2.3.4+3.4.3+...+99.100.3\) 

\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)....99.100.\left(101-98\right)\) 

\(3A=\left(1.2.3+2.3.4+3.4.5+...+99.100.101\right)-\left(0.1.2+1.2.3+2.3.4+...+98.99.100\right)\)

\(3A=99.100.101-0.1.2\) 

\(3A=999900-0\)

\(3A=999900\)

\(A=999900:3\)

\(\Rightarrow A=333300\)

a) \(\left(2x+1\right)^3=125\)

\(\left(2x+1\right)^3=5^3\)

\(2x+1=5\)

\(2x=5-1\) 

\(2x=4\)

\(x=4:2\)

\(\Rightarrow x=2\)

b) \(5^{x+1}=625\)

\(5^{x+1}=5^4\)

\(x+1=4\)

\(x=4-1\)

\(\Rightarrow x=3\)

a-b=(1.2+2.3+3.4+4.5+...+98.99)-(1²+2²+3²+...+98²)

1.2+2.3+3.4+4.5+...+98.99-1²-2²-3²-...-98²

^{=1(2-1)+2(3-2)+...+98(99-98)}

^=1+2+...+98

Đến đây bạn tự tính

b=1²+2²+3²+............+98²

b=1.(2-1)+2.(3-1)+3.(4-1)+..............+98.(99-1)

b=1.2-1+2.3-2+3.4-3+.............+98.99-98

b=(1.2+2.3+3.4+..................+98.99)-(1+2+3+............+98)

a-b=1+2+3+...............+98

a-b=\(\frac{98.\left(98+1\right)}{2}\)

a-b=4851

                                   Vậy A-B=4851

\(\text{a=1.2+2.3+3.4+......+98.99}\)

\(=1\left(1\text{+}1\right)\text{+}2\left(2\text{+}1\right)\text{+}3\left(3\text{+}1\right)\text{+}.........\text{+}98\left(98\text{+}1\right)\)

\(=1^2\text{+}1\text{+}1^2\text{+}2\text{+}3^2\text{+}3\text{+}...\text{+}98^2\text{+}98\)

\(=b\text{+}\left(1\text{+}2\text{+}3\text{+}...\text{+}98\right)\)

\(=b\text{+}\left(98\text{+}1\right).98:2\)

\(=b\text{+}4851\)

\(\Rightarrow a-b=4851\)

dễ ợt

Ta có :\(B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}.....\frac{98^2}{98.99}=\frac{\left(1.2.3.4...98\right).\left(1.2.3.4...98\right)}{\left(1.2.3.4...98\right).\left(2.3.4.5...99\right)}=\frac{1}{99}\)

Lại có A = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)

Lại có \(A:B=\frac{98}{99}:\frac{1}{99}=98\)

=> A = 98B

các bạn có  về sweet home

Áp dụng công thức ta có :

\(A=1.2+2.3+3.4+...+99.100=\frac{99.100.101}{3}=333300\)

A=1.2+2.3+3.4+4.5+.....+98.99+99.100 Rút gọn đi ta còn:

A=1+100

=>A=101

Ta có : A = 1.2 + 2.3 + 3.4 + ....... + 99.100

=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 99.100.101

=> 3A = 99.100.101

=> A = 99.100.101/3

=> A = 333300