Sửa đề: \(-1+3-5+7-...-97+99\)

1) Ta có: \(-1+3-5+7-...-97+99\)

\(=\left(-1+3\right)+\left(-5+7\right)+...+\left(-97+99\right)\)

\(=2+2+...+2=2\cdot50=100\)

2) Ta có: \(1+2-3-4+...+97+98-99-100\)

\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(97+98-99-100\right)\)

\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)

\(=\left(-4\right)\cdot25=-100\)

a: \(A=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)

\(=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+\dfrac{100}{100}\)

\(=100\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)=100B

=>B/A=1/100

b: \(A=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\left(1\right)\)

\(=\dfrac{50}{49}+\dfrac{50}{48}+....+\dfrac{50}{2}+\dfrac{50}{50}\)

\(=50\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)

\(B=\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{49}+\dfrac{2}{50}\)

\(=2\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)

=>A/B=25

1-2+3-4...+99-100

= (1-2)+(3-4)...+(99-100)

= -1.(100/2)

= -50

Cho mik hỏi cách làm bài này

Tính nhanh 1 1/2x1 1/3 × 11/4×...x 1 1/99×1 1/100

`S=1+2+3+4+5+...+99+100`

`S=((100+1)[(100-1):1+1])/(2)=5050`

Số số hạng là:

( 100 - 1 ) : 1 + 1 = 100 ( số )

Tổng của S trên là:

( 100 + 1 ) x 100 : 2 = 5050

Đáp số: 5050

S=1+2+3+4+5+...+99+100

SSH:(100-1):1+1=100(SSH)

Tổng là:(100+1)x100:2=5050

\(=\left(-1\right)^{1+2+3+...+100}=\left(-1\right)^{5050}=1\)

Xin hãy giúp mình

a,=3/2*4/3*....100/99

=3*4*5*....*100/2*3*...*99

=100/2=50

b, nhân lên băng:

1*2*3*...*99/2*3*...*100=1/100

\(D=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\)

\(E=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{98\cdot99}\right)\)

\(=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{99}-1+\dfrac{1}{99}=\dfrac{2}{99}-1=-\dfrac{97}{99}\)