Question

1) 

a) tim tat ca cac chu so a,b,c thoa man : abc - cba = 6b3

b) tim x biet : 1/21 + 1/28 +1/36 + ........ + 2/x(x+1) = 2/9

Answer 1

abc-cba=a.100+b.10+c-c.100-b.10-a

=99(a-c)=6b3

=> 6b3 chia hết cho 99=>b=9

=> a-c=693:99=7=>(a,c) E {(8;1);(9;2)}

Vậy (a,b,c) E {(8;9;1);(9;9;2)}

Answer 2

\(abc-cba=6b3\)

\(\Leftrightarrow100a+10b+c-100c-10b-a=600+10b+3\)

\(\Leftrightarrow99a-99c=603+10b\)

\(\Leftrightarrow99\left(a-c\right)=603+10b\)

\(\Leftrightarrow\hept{\begin{cases}a=8\\b=9\\c=1\end{cases}}\)

Answer 3

2/\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Leftrightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Leftrightarrow2\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)

\(\Leftrightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Leftrightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\Leftrightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{18}\Leftrightarrow x+1=18\Leftrightarrow x=17\)