\(3\times10^3+2\times10^2+5\times10^1\\ =3\times1000+2\times100+5\times10\\ =3000+200+50\\ =3250\)

`@` `\text {Ans}`

`\downarrow`

\(3\cdot10^3+2\cdot10^2+5\cdot10^1\)

\(=10\cdot\left(3\cdot10^2+2\cdot10+5\right)\)

\(=10\cdot\left[10\left(3\cdot10+2\right)+5\right]\)

\(=10\cdot\left[10\left(30+2\right)+5\right]\)

\(=10\cdot\left(320+5\right)\)
\(=10\cdot325\)

\(=3250\)

Ta có: \(C=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)

\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+\dfrac{1}{122}+\dfrac{1}{123}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+\dfrac{1}{182}+\dfrac{1}{183}+...+\dfrac{1}{200}\right)\)

\(\Leftrightarrow C>20\cdot\dfrac{1}{120}+30\cdot\dfrac{1}{150}+30\cdot\dfrac{1}{180}+20\cdot\dfrac{1}{200}\)

\(\Leftrightarrow C>\dfrac{1}{6}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{19}{30}=\dfrac{76}{120}\)

\(\Leftrightarrow C>\dfrac{75}{120}=\dfrac{5}{8}\)

hay \(C>\dfrac{5}{8}\)(đpcm)

j mà  nhìu zu zậy làm bao giờ mới xong

Ủng hộ mk đi các bạn
 

A= [(1+101)x101:2]-(102-103)
A= 5151+1
A=5152

B= [1+(-3)]+[4+(-5)]+.......[101+(-103)]+105
B= (-2)+(-2)...........+(-2)+105

=> A>B
B=(-2)x26+105
B=(-56)+105
B= 49

cái => A>B nó nằm ở dưới cùng ấy. Nãy gõ chứ nó bị nhảy phím