= 2-1/1.2 + 3-2/2.3 + 4-3/3.4 + ...... + 3024-3023/3023.3024

= 1-1/2+1/2-1/3+1/3-1/4+.....+1/3023-1/3024

= 1- 1/3024 = 3023/3024

=1-1/2+1/2-1/3+1/3-1/4+.......+1/3023-1/3014

=1-1/3024=3023/3024

k cho mình nha

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{3023.3024}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{3023}-\frac{1}{3024}\)

\(=1-\frac{1}{3024}\)

\(=\frac{3023}{3024}\)

\(3.\dfrac{4}{10}< 3.\dfrac{9}{10}\\ 5.\dfrac{1}{10}>2.\dfrac{9}{10}\\ 3.\dfrac{4}{10}=3.\dfrac{2}{5}\)

a) \(3\cdot\dfrac{4}{10}< 3\cdot\dfrac{9}{10}\)

b) \(5\cdot\dfrac{1}{10}< 2\cdot\dfrac{9}{10}\)

c) \(3\cdot\dfrac{4}{10}=3\cdot\dfrac{2}{5}\)

8⁵=(2³)⁵=2¹⁵=2.2¹⁴

3.4⁷=3.(2²)¹⁴=3.2¹⁴

vì 2<3 nên 2.2¹⁴<3.2¹⁴

hay 8⁵<3.4⁷

3³⁵=3².3³³=9.(3³)¹¹=9.27¹¹

5²³=5¹.5²²=5.(5²)¹¹=5.25¹¹

vì 9>5 và 27>25 nên 9.27¹¹>5.25¹¹

hay 3³⁵>5²³

xem lại câu này

2¹⁶=2³.2¹³=8.2¹³

vì 7<8 nên 7.2¹³<8.2¹³

hay 7.2¹³<2¹⁶

Đặt A = 1.2 + 2.3 + 3.4 + ...... + 99.100

3A= 3.(1.2 + 2.3 + 3.4 + ..... +99.100)

3A=1.2.(3-0) + 2.3.(4-1) +.....+99.100.(101-98)

 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=333300

đặt A = 1.2 + 3.4 + 4.5 +...+ 99.100

A=1.2+2.3+3.4+4.5+...+99.100

=>3A=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3

=1.2.3+2.3.﴾4‐1﴿+3.4.﴾5‐2﴿+4.5.﴾6‐3﴿+...+99.100.﴾101‐98﴿

=1.2.3+2.3.4‐1.2.3+3.4.5‐2.3.4+4.5.6‐3.4.5+...+99.100.101‐98.99.100

=1.2.3‐1.2.3+2.3.4‐2.3.4+3.4.5‐3.4.5+4.5.6‐4.5.6+...+99.100.101

=99.100.101=999900

=>A=999900:3=333300

Vậy A=333300

đặt A = 1.2 + 3.4 + 4.5 +...+ 99.100

A=1.2+2.3+3.4+4.5+...+99.100

=>3A=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3

=1.2.3+2.3.﴾4‐1﴿+3.4.﴾5‐2﴿+4.5.﴾6‐3﴿+...+99.100.﴾101‐98﴿

=1.2.3+2.3.4‐1.2.3+3.4.5‐2.3.4+4.5.6‐3.4.5+...+99.100.101‐98.99.100

=1.2.3‐1.2.3+2.3.4‐2.3.4+3.4.5‐3.4.5+4.5.6‐4.5.6+...+99.100.101

=99.100.101=999900

=>A=999900:3=333300

Vậy A=333300

1230

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^_^

D=1.2+2.3+3.4+...+19.20

=>3D=1.2.3+2.3.3+3.4.3+...+19.20

=1.2.3+2.3(4-1)+3.4(5-2)+...+19.20(21-18)

=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+19.20.21-18.19.20

=>3D=1.2.3+2.3.3+3.4.3+...+19.20

=1.2.3+2.3(4-1)+3.4(5-2)+...+19.20(21-18)

=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+19.20.21-18.19.20

=19.20.21=7980

=>D=7980:3=2660

Vậy D=2660

Ta có : \(-\left|x+1,2\right|\le0\forall x\)

Suy ra : \(A=-\left|x+1,2\right|+3,4\le3,4\forall x\)

Vậy \(A_{min}=3,4\) khi \(x=-1,2\)

Sorry bạn nhé bài đầu tiên bạn sửa chỗ min thành "max" nhé !

Ta có : \(\left|x+1,2\right|\ge0\forall x\)

Suy ra : B = \(\left|x+1,2\right|-3,4\ge-3,4\forall x\)

Vậy B_min = -3,4 khi x = -1,2

Ta có : A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)

\(\Rightarrow\)3A = 1.2.(3-0)+2.3.(4-1)+3.4.(5-2).....n.(n+1).[(n+2)-(n-1)]

\(\Rightarrow\)3A= 1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+....+n.(n+1)(n+2)-(n-1)n(n+1)

\(\Rightarrow\)3A= (1.2.3-1.2.3)+(2.3.4-2.3.4)+....+[(n-1).n.(n+1)-(n-1)n(n+1)]+n.(n+1)(n+2)

\(\Rightarrow\)3A=n.(n+1)(n+2)

\(\Rightarrow\)A=\(\frac{\text{n.(n+1)(n+2)}}{3}\)

Tại sao có 3A