quen lắm

100+x=200+300

100+x=500

       x=500-100

       x=400

\(100+x=200+300\)

\(100+x=500\)

              \(x=500-100\)

              \(x=400\)

1/100¹=1/100

1/100²=1/99.100

...........................

1/200¹=1/200

=>A<1/100+1/99.100+.....+1/200

=>A<1/100+1/99+1/100+1/200

=>A<1/99+1/200

Vậy A<1//99+1/200

bạn ko cho biết làm gì nen mình so sanh no nha 

mong ban chon minh

\(^{1^{100}}\)+ \(^{1^{200}}\)= 100 + 200

                      = 300

\(1^{100}\) \(+\) \(1^{200}\)

\(=\) \(1\) \(+\) \(1\)

\(=\)\(2\)

a)x=30

b)x=65

bn giải cụ thể hơn đi , mình ko hiểu

=>x(x+1)=1000.1001

Mà x<x+1 là 1 đơn vị

1000<1001 1 đơn vị

=>x=1000

phân tích 1001000 ra thừa số nguyên tố rồi nhóm sao cho ổn

x = 50

k mk,mk k lai bn

50 nhe co ai ket ban voi minh ko

 -100/200+25/75+17/23+-6/36

= -1/2+1/3+17/23+-1/6

= (-1/2+1/3+-1/6)+17/23

= -1/3+17/23

= 28/69

cfghhg

a, Ta có: \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

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

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

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

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

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

b, Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)(đpcm)

a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...\)\(+\frac{99}{100}\)
Xét: \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
    = \(\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)
    = \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)\)                                                          
    = \(\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)( có 99 số hạng là 1 )
    = \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
    = \(\left(99+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
    = \(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrow100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)( đpcm )
Vậy: ...