\(\frac{B}{A}=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

\(\frac{B}{A}=\frac{1+\left[\frac{1}{99}+1\right]+\left[\frac{2}{98}+1\right]+\left[\frac{3}{97}+1\right]+...+\left[\frac{98}{2}+1\right]}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

\(\frac{B}{A}=\frac{\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

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

Vậy : \(\frac{B}{A}=100\)

Ta có:

\(B=\frac{1}{99}+\frac{2}{98}+...+\frac{99}{1}\)

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

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

\(=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)

\(=100.A\)

\(\Rightarrow\frac{B}{A}=100\)

l,¬p,p¬m[p,¬p,¬

  1+2-3-4+5+6-7-8+..........+97+98-99-100

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

=(-4)+(-4)+.....+(-4) (25 số -4)

=(-4)x25

=-100

Ta có: \(\frac{2^{100}+2^{101}+2^{102}}{2^{97}+2^{98}+2^{99}}=\frac{2^{100}\left(1+2+2^2\right)}{2^{97}\left(1+2+2^2\right)}=2^3=8\)

a Ta có 

B= 1-2-3+4-5-6-7+8......+ 97 -98-99+100

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

=       0 +0+... +0 (25 cs 0)

=0 x25=0

a)B=0 

25 cs 0 là gì vậy ?

A=-1+(-1)+...+(-1) {có 50 số hạng}

=-1.50=-50

A=1-2+3-4+...+99-100       SSH=(100-1):1+1=100 Sh

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

vì chia thành cặp suy ra 100:2 =50 cặp

A=(-1)+(-1)+...(-1)

A=(-1).50

A=-50

bạn giúp mình trả lời chi tiết câu c nha cám ơn các ban

ta có : \(100^2-99^2+98^2-97^2+...+2^2-1^2\)

\(=\left(100^2-1^2\right)-\left(99^2-2^2\right)+\left(98^2-3^2\right)-...+\left(52^2-49^2\right)-\left(51^2-50^2\right)\)

\(=101\left(100-1\right)-101\left(99-2\right)+101\left(98-3\right)-...+101\left(52-49\right)-101\left(51-50\right)\)

\(=101.99-101.97+101.95-...+101.3-101.1\)

\(=101\left(99-97+95-93+...+3-1\right)\)

\(=101.\left(2+2+2+...+2\right)=101.2.25=5050\)

Mình có cách khác nha !

\(100^2-99^2+98^2-97^2+...+2^2-1^2\)

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

\(=100+99+98+97+...+2+1\)

\(=\dfrac{100.101}{2}=5050\)

\(100^2-99^2+98^2-97^2+...+2^2-1^2\)

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

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

\(=100+99+98+97+...+2+1\)

\(=\dfrac{\left(100+1\right).100}{2}=\dfrac{101.100}{2}=\dfrac{10100}{2}=5050\)