\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

<=>  \(\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)

<=>  \(\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)

<=>  \(\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

Nhận thấy:   \(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\)

=>  \(x-105=0\)

<=>  \(x=105\)

\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

\(\Leftrightarrow\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}-\frac{x-100}{5}-\frac{x-101}{4}-\frac{x-102}{3}=0\)

\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)-\left(\frac{x-100}{5}-1\right)-\left(\frac{x-101}{4}-1\right)-\left(\frac{x-102}{3}-1\right)=0\)

\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}-\frac{x-105}{5}-\frac{x-105}{4}-\frac{x-105}{3}=0\)

\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

\(\Leftrightarrow x-105=0\left(Vì\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)\)

\(\Leftrightarrow x=105\)

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

`@` `\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