1/

6 = 1*2*3

24 = 2*3*4

.......

Số thứ 100: 100*101*102 

TỔng dãy trên là A thì bằng:

A = 1*2*3 + 2*3*4 + ..... + 100*101*102

4A = 1*2*3*4 + 2*3*4*4 + .... + 100*101*102*4

4A = [1*2*3*4 - 0*1*2*3]+ [2*3*4*5 - 1*2*3*4]+ ...+[100*101*102*103 - 99*100*101*102]

4A = 0*1*2*3 + [1*2*3*4-1*2*3*4]+[2*3*4*5-2*3*4*5]+..........+[99*100*101*101-99*100*101*102] + 100*101*102*103

4A = 100*101*102*103

A = 25*101*102*103 = 26527650

2/

\(A=\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{73\cdot76}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{73}-\frac{1}{76}\)

\(A=\frac{1}{4}-\frac{1}{76}=\frac{9}{38}\)

P/s: Vì tử bằng khoẳng cách dưới mẫu nên ta có thể rút gọn nhanh như vậy

2)    \(A=\frac{3}{4.7}+\frac{3}{7.10}+........+\frac{3}{73.76}\)

       \(A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.......+\frac{1}{73}-\frac{1}{76}\)

       \(A=\frac{1}{3}-\frac{1}{76}\)

      \(A=\frac{73}{228}\)

a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)

\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)

b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)

Ta có :

\(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{9}{10!}\)

\(A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{10-1}{10!}\)

\(A=\left(\frac{2}{2!}-\frac{1}{2!}\right)+\left(\frac{3}{3!}-\frac{1}{3!}\right)+\left(\frac{4}{4!}-\frac{1}{4!}\right)+...+\left(\frac{10}{10!}-\frac{1}{10!}\right)\)

\(A=\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+...+\left(\frac{1}{9!}-\frac{1}{10!}\right)\)

\(A=1-\frac{1}{10!}< 1\)

vậy A < 1 vì \(0< \frac{1}{10!}< 1\)

CHTT là sao

Áp dụng tính chất (a - b)(a + b) = a² + ab - ab - b² = a² - b²

Ta có : \(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{19}{\left(9.10\right)^2}\)

\(=\frac{1}{1.2}.\frac{3}{1.2}+\frac{1}{2.3}.\frac{5}{2.3}+...+\frac{1}{9.10}.\frac{19}{9.10}\)

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

\(=1^2-\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2-\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{9}\right)^2-\left(\frac{1}{10}\right)^2=1^2-\left(\frac{1}{10}\right)^2=1-\frac{1}{100}=\frac{99}{100}< 1\)

2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)

              \(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)

Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)

Vậy \(2^{332}< 3^{223}\)

1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)

\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)

Nên suy ra \(10A>10B\Rightarrow A>B\)

`3/(-10) ; 1/(-2) ; 4/(-5)=> -3/10 ; -1/2 ; -4/5`

ta có : `-1/2=(-1xx5)/(2xx5)=-5/10 ; -4/5=(-4xx2)/(5xx2)=-8/10`

vậy `3/(-10) < 1/(-2) < 4/(-5)`

`--------------------`

`2/(-10) ; 7/(-5) ; -1/2=>-2/10 ;-7/5;-1/2`

ta có : `-7/5=(-7xx2)/(5xx2)=-14/10; -1/2=(-1xx5)/(2xx5)=-5/10`

vậy `2/(-10) < -1/2 < 7/(-5)`

`---------------------`

`7/(-4) ; -2/5 ; -3/10=> -7/4;-2/5;-3/10`

ta có : `-7/4=(-7xx5)/(4xx5)=-35/20 ; -2/5=(-2xx4)/(5xx4)=-8/20;-3/10=(-3xx2)/(10xx2)=-6/20`

vậy 7/(-4) > -2/5 > -3/10`

GIÚP MIK VỚI ĐG CẦN GẤP