A = 10/1x2 + 10/2x3 + 10/3x4 +........ + 10/98x99 + 10/99x100

A=10x(1/1x2+1/2x3+1/3x4+..+1/98x99+1/99x100)

A= 10x( 1-1/2+1/2-1/3+1/3-1/4+..+1/99-1/100)

A=10x( 1-1/100)

A=10x99/100

A=99/10

A = đề

A=10x(1/1x2+1/2x3+1/3x4+..+1/98x99+1/99x100)

A= 10x( 1-1/2+1/2-1/3+1/3-1/4+..+1/99-1/100)

A=10x( 1-1/100)

A=10x99/100

A=99/10

 A = 1×2 + 2×3 + 3×4 + ... + 98×99

3A = 1×2×(3-0) + 2×3×(4-1) + 3×4×(5-2) + ... + 98×99×(100-97)

3A = 1×2×3 - 0×1×2 + 2×3×4 - 1×2×3 + 3×4×5 - 2×3×4 + ... + 98×99×100 - 97×98×99

3A = 98×99×100

A = 98×33×100

A = 323400

2) Áp dụng a/b < 1 => a/b < a+m/b+m (a,b,m thuộc N*)

Ta có:

10²⁰¹² + 1/10²⁰¹³ + 1 < 10²⁰¹² + 1 + 9/10²⁰¹³ + 1 + 9

                                        < 10²⁰¹² + 10/10²⁰¹³ + 10    

                                        < 10.(10²⁰¹¹ + 1)/10.(10²⁰¹² + 1)

                                        < 10²⁰¹¹ + 1/10²⁰¹² + 1

Vào lúc: 2016-07-17 13:22:30 Xem câu hỏi

1) A = 1×2 + 2×3 + 3×4 + ... + 98×99

3A = 1×2×(3-0) + 2×3×(4-1) + 3×4×(5-2) + ... + 98×99×(100-97)

3A = 1×2×3 - 0×1×2 + 2×3×4 - 1×2×3 + 3×4×5 - 2×3×4 + ... + 98×99×100 - 97×98×99

3A = 98×99×100

A = 98×33×100

A = 323400

2) Áp dụng a/b < 1 => a/b < a+m/b+m (a,b,m thuộc N*)

Ta có:

10²⁰¹² + 1/10²⁰¹³ + 1 < 10²⁰¹² + 1 + 9/10²⁰¹³ + 1 + 9

                                        < 10²⁰¹² + 10/10²⁰¹³ + 10    

                                        < 10.(10²⁰¹¹ + 1)/10.(10²⁰¹² + 1)

                                        < 10²⁰¹¹ + 1/10²⁰¹² + 1

=1x2x3+2x3x3+...+98x99x3

=1x2x3+2x3x(4-1)+...+98x99x(100-97)

=1x2x3+2x3x4-1x2x3+...+98x99x100-97x98x99

=98x99x100

=970200

\(\frac{1}{1x2}+\frac{1}{1x3}+...+\frac{1}{999x1000}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)

\(=1-\frac{1}{1000}\)

\(=\frac{999}{1000}\)

1/1x2+1/2x3+1/3x4+...+1/999x1000

=1-1/2+1/2-1/3+1/3-1/4+...+1/999-1/1000

=1-1/1000

=1000/1000-1/1000

=999/1000

999 / 1000

\(4\frac{1}{10}+4\frac{2}{10}+4\frac{3}{10}+4\frac{4}{10}\)

= \((4+4+4+4)+(\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10})\)

= \(4\times4+\frac{1}{10}(1+2+3+4)\)

= \(16+\frac{1}{10}.10\)

= \(16+1=17\)

\(4\frac{1}{10}+4\frac{2}{10}+4\frac{3}{10}+4\frac{4}{10}\)

\(=4+\frac{1}{10}+4+\frac{2}{10}+4+\frac{3}{10}+4+\frac{4}{10}\)

\(=\left(4+4+4+4\right)+\left(\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}\right)\)

\(=16+\frac{10}{10}=16+1=17\)

\(A=\frac{1.98+2.97+3.96+...+98.1}{1.2+2.3+3.4+...+98.99}=\frac{1.\left(100-2\right)+2\left(100-3\right)+3\left(100-4\right)+...+98\left(100-99\right)}{1.2+2.3+3.4+...+98.99}\)

\(A=\frac{1.100-1.2+2.100-2.3+3.100-3.4+...+98.100-98.99}{1.2+2.3+3.4+...+98.99}\)

\(A=\frac{\left(1.100+2.100+3.100+...+98.100\right)-\left(1.2+2.3+3.4+...+98.99\right)}{1.2+2.3+3.4+...+98.99}\)

\(A=\frac{100\left(1+2+3+...+98\right)}{1.2+2.3+3.4+...+98.99}-1\)

Ta có: 1+2+3+...+98=98.99:2=4851

Đặt B=1.2+2.3+3.4+...+98.99  => 3B=1.2.3+2.3.3+3.4.3+...+98.99.3 = 1.2.3+2.3.(4-1)+3.4(5-2)+...+98.99(100-97)

=> 3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.99 = 98.99.100

=> B=33.98.100. Thay vào A được:

\(A=\frac{100.4851}{33.98.100}-1=\frac{3}{2}-1=\frac{1}{2}\)

=5(x1/1x2 + 1/2x3 +... +1/99x100)

= 5 x( 1/1 - 1/2 +1/2 -1/3 +... +1/99 -1/100)

= 5 x( 1 /1- 1/100)

= 5 x99/100 

= 99/ 20