=0:{2+4+6+...98}=0

=[1+3+5+7+...97+99]x[45x3-45x3]

=[----------------------------]x0=0

Dấu gạch trên là gì đấy?

a, [ 0 x 1 x 2 x 3 ...x 99 x 100] : [2 + 4 + 6 + ... 98]

Vì có chữ số 0 mà 0 nhân số nào cũng bằng 0 

=> 0 : ( 2 + 4 + 6 + ... 98 )

Vì số nào chia 0 cũng bằng 0 

=> 0 : ( 2 + 4 + 6 +.. + 98 ) = 0

b, Đặt A = 1 + 3 + 5 + 7 + ... + 97 + 99 )

    Đặt B = 45x 3 - 45 x 2 - 45

B = 45 x 3 - 45 x 2 - 45

B = 45 x 3 - 45 x 2 - 45 x 1

B = 45 x ( 3 - 2 - 1 )

B = 45 x 0

B = 0

Vì 0 nhân số nào cũng = 0 

=> ( 1 + 3 + 5 + 7 + ... + 97 +99 ) x 0 = 0

c, Bạn chỉ cần biến đổi tử số hoặc mẫu số giống nhau thì kết quả sẽ = 1 nha

( 0 x 1 x 2 x 3 x .... x 99 x 100 ) : ( 2 + 4 + 6 + ... + 98 )

= 0 : ( 2 + 4 + 6 + ... + 98 )

= 0

không biết giải

2001 

____

1991

1/1*2+1/2*3+1/3*4+1/4*5+...1/99+100

Đặt A=1+2+2²+2³+...+2¹⁰⁰

suy ra 2A=2+2²+2³+...+2¹⁰⁰

suy ra 2A-A=(2+2²+2³+...+2¹⁰¹)-(1+2+2²+2³+...+2¹⁰⁰⁾

                 =2¹⁰¹-1

Vậy 1+2+2²+2³+...+2¹⁰⁰=2¹⁰¹-1

 A=1+2+2^2+2^3+...+2^100

2A=2+2²+2³+2⁴+...+2¹⁰¹

2A-A=2¹⁰¹-1

Vậy A= 2¹⁰¹-1

1) x=100

\(S=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\\ 3S=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot3\cdot4+...+3\cdot99\cdot100\\ 3S=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\\ 3S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+....+99\cdot100\cdot101-98\cdot99\cdot100\\ 3S=99\cdot100\cdot101\\ S=\dfrac{99\cdot100\cdot101}{3}=33\cdot100\cdot101=3300\cdot101=333300\)

      A=100/1 x 2 + 100/2 x 3 + 100/3 x 4 +...+100/99 x 100

A/100=1/1 x 2 + 1/2 x 3 + 1/3 x 4 +...+1/99 x 100

A/100=2-1/1x2 + 3-2/2x3 + ... + 100-99/99x100

A/100=1-1/2 + 1/2-1/3+...+1/99-1/100

A/100=1-1/100

A/100=99/100

A=99/100x100=99

Vậy A=99.

Ta có:

\(\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+...+\frac{100}{99.100}\)

\(\Rightarrow100.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(\Rightarrow100.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow100.\left(\frac{1}{1}-\frac{1}{100}\right)\Leftrightarrow100.\frac{99}{100}=99\)

\(\text{Ta có :}\)

\(\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+...+\frac{100}{99.100}\)

\(=100.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

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

a) Số số hạng: \(\frac{\left(99-1\right)}{1}+1=99\)

Tổng: \(\frac{99+1}{2}\cdot99=4950\)

b) Số số hạng: \(\frac{\left(100-2\right)}{2}+1=50\)

Tổng: \(\frac{100+2}{2}\cdot50=2550\)

c) \(S=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)

\(3\cdot S=1\cdot2\left(3-0\right)+2\cdot3\left(4-1\right)+3\cdot4\left(5-2\right)+...+99\cdot100\left(101-98\right)\)

\(3\cdot S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+99\cdot100\cdot101-98\cdot99\cdot100\)

\(3\cdot S=99\cdot100\cdot101\)

Vậy, \(S=\frac{1}{3}\cdot99\cdot100\cdot101=333300\)

= 1/1x2 + 1/2x3 + 1/3x4 ...... +1/9x10

= 1-1/2+1/2-1/3+1/3-1/4+........+1/9-1/10

=1-1/10=9/10

đặt A=1/1 x 1/2 + 1/2 x 1/3 + 1/3 + 1/4 + .......... + 1/9 x 1/10

\(A=\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+...+\frac{1}{9}\cdot\frac{1}{10}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{9}{10}\)

đặt B=2/1 x 2 + 2/2 x 3 + 2/3 x4 + .............. + 2/98 x 99 + 2/99 x 100

\(B=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

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

\(=2\left(1-\frac{1}{100}\right)\)

\(=2\times\frac{99}{100}\)

\(=\frac{99}{50}\)

S = 1 x 2 + 2 x 3 + ... + 99 x 100

3S = 1 x 2 x 3 + 2 x 3 x (4 - 1) + ..... + 99 x 100 x (101 - 98)

3S = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + .... + 99 x 100 x 101 - 98 x 99 x 100

3S = 99 x 100 x 101 = 999900

S = 999900 : 3 = 333300 

999900

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