bằng 1

Ta có:

$A=\dfrac{1+(1+2)+(1+2+3)+\cdots+(1+2+\cdots+98)}{1\times98+2\times97+3\times96+\cdots+98\times1}$

Tử số:

$1+(1+2)+(1+2+3)+\cdots+(1+2+\cdots+98)$

$=1+2+2^2+?$

Cách dễ nhất là viết:

$1+2+\cdots+k=\dfrac{k(k+1)}2$

Nên tử số:

$=\dfrac{1\cdot2}{2}+\dfrac{2\cdot3}{2}+\cdots+\dfrac{98\cdot99}{2}$

$=\dfrac{1}{2}(1\cdot2+2\cdot3+\cdots+98\cdot99)$

$=\dfrac{1}{2}\cdot\dfrac{98\cdot99\cdot100}{3}$

$=161700$

Mẫu số:

$1\cdot98+2\cdot97+3\cdot96+\cdots+98\cdot1$

$=1(99-1)+2(99-2)+\cdots+98(99-98)$

$=99(1+2+\cdots+98)-(1^2+2^2+\cdots+98^2)$

$=99\cdot\dfrac{98\cdot99}{2}-\dfrac{98\cdot99\cdot197}{6}$

$=161700$

Do đó:

$A=\dfrac{161700}{161700}=1$

Vậy $A=1$.

a) \(100-99+98-97+...+4-3+2-1=\)

\(\left(100-99\right)+\left(98-97\right)+...+\left(4-3\right)+\left(2-1\right)=\)

\(1+1+1+...+1+1\left(50con1\right)=50\)

b) Ta xen các số lẻ vào các số chẵn :

\(100+98-97+96-95+...+2-1=\)

\(100+\left(98-97\right)+\left(96-95\right)+...+\left(2-1\right)=\)

\(100+1+1+1+...+1\left(49con1\right)=149\)

Ủng hộ mik nha

a)=-100

b)=101

\(A=\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+\frac{4}{96}+...+\frac{98}{2}+\frac{99}{1}\)

\(A=1+\left(\frac{1}{99}+1\right)+\left(\frac{2}{98}+1\right)+\left(\frac{3}{97}+1\right)+\left(\frac{4}{96}+1\right)+...+\left(\frac{98}{2}+1\right)\)

\(A=\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+\frac{100}{96}+...+\frac{100}{2}\)

\(A=100.\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{100\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=100\)

tính B

100 - 99 + 98 - 97 + 96 - 95 + ... + 2 - 1

= (100 - 99) + (98 - 97) + (96 - 95) + ... + (2 - 1)

= 1+1+1+1+...+1+1 ( có 50 cặp số 1 ) 

= 1 x 50

=50

100 + 98 + 96 + ... + 2 - 97 - 95 - ... - 1

= 100 + (98 - 97) + (96 - 95) + ... + (2 - 1)

= 100 + 1 + 1 + 1 +...+ 1 ( có 49 số 1 )

= 100 + 1 x 49

= 100 + 49

= 149