\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

Ta có: 

5¹⁹⁹²  = ( 5⁴ )⁴⁹⁸ = 625⁴⁹⁸ = ( ..... 0625)

Vậy 4 chữ số tận cùng của số 5¹⁹⁹² là 0625.

\(a)\) \(\left|2x-1\right|+5=12-x\)

\(\Leftrightarrow\)\(\left|2x-1\right|=12-x-5\)

\(\Leftrightarrow\)\(\left|2x-1\right|=7-x\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-1=7-x\\2x-1=x-7\end{cases}\Leftrightarrow\orbr{\begin{cases}2x+x=7+1\\2x-x=-7+1\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=8\\x=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=-6\end{cases}}}\)

Vậy \(x=\frac{8}{3}\) và \(x=-6\)

\(b)\) \(\left|2x+1\right|=x-1\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}2x+1=x-1\\2x+1=1-x\end{cases}\Leftrightarrow\orbr{\begin{cases}2x-x=-1-1\\2x+x=1-1\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-2\\3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=0\end{cases}}}\)

Vậy \(x=-2\) hoặc \(x=0\)

Chúc bạn học tốt ~

\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)

\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}-\frac{28}{41}\)

\(=\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)+\frac{1}{3}\)

\(=\left(-1\right)+1+\frac{1}{3}\)

\(=0+\frac{1}{3}\)

\(=\frac{1}{3}\)

=1/3+12/67+13/41-79/67+28/41

=1/3+(12/67-79/67)+(13/41+28/41)

=1/3+(-1)+1

=1/3+0

=1/3

a) \(\frac{4}{9}.\left(\frac{5}{7}+\frac{2}{5}\right).\frac{4}{9}.\frac{-2}{7}+\frac{7}{5}\)

= \(\frac{4}{9}.\left(\frac{25}{35}+\frac{14}{35}\right).\frac{-8}{63}+\frac{7}{5}\)

= \(\frac{4}{9}.\frac{39}{35}.\frac{-8}{63}+\frac{7}{5}\)

= \(\frac{4.39.\left(-8\right)}{9.35.63}+\frac{7}{5}\)

= \(\frac{-416}{6615}+\frac{7}{5}\)

= \(\frac{-416}{6615}+\frac{9261}{6615}\)

= \(\frac{8845}{6615}\)

= \(\frac{1769}{1323}\)

Theo đề bài ta có:
\(\frac{561}{143}< x\frac{12}{13}< \frac{1463}{247}\)

\(\Leftrightarrow\frac{51}{13}< \frac{13x+12}{13}< \frac{77}{13}\)

\(\Leftrightarrow51< 13x+12< 77\)

\(\Leftrightarrow39< 13x< 65\)

\(\Leftrightarrow x=4\)

Vậy x = 4

Ta có : 

\(\frac{561}{143}< x\frac{12}{13}< \frac{1463}{247}\)

\(\Leftrightarrow\)\(\frac{51}{13}< \frac{13x+12}{13}< \frac{77}{13}\)

\(\Leftrightarrow\)\(51< 13x+12< 77\)

\(\Leftrightarrow\)\(39< 13x< 65\)

Mà x là phần nguyên nên \(13x\inℤ\)

\(\Rightarrow\)\(13x=52\)

\(\Rightarrow\)\(x=\frac{52}{13}\)

\(\Rightarrow\)\(x=4\)

Vậy \(x=4\) hay hỗn số cần tìm là \(4\frac{12}{13}\)

Chúc bạn học tốt ~

\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+.........+\frac{1}{24.25}\)

= \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-........-\frac{1}{24}+\frac{1}{24}-\frac{1}{25}\)

= \(\frac{1}{5}-\frac{1}{25}\)

= \(\frac{5}{25}-\frac{1}{25}\)

= \(\frac{4}{25}\)

Ta có : 

\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

\(=\)\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

\(=\)\(\frac{1}{5}-\frac{1}{25}\)

\(=\)\(\frac{5}{25}-\frac{1}{25}\)

\(=\)\(\frac{4}{25}\)

Vậy \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}=\frac{4}{25}\)

Chúc bạn học tốt ~

Ta có : 

\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)

Vì \(a+b⋮6\)

\(\Rightarrow\)\(\left(a+b\right)^3⋮6\)

\(\Rightarrow\)\(3ab\left(a+b\right)⋮6\)

\(\Rightarrow\)\(\left(a+b\right)^3-3ab\left(a+b\right)⋮6\) ( đpcm ) 

Vậy \(a^3+b^3⋮6\)

Chúc bạn học tốt ~

Có ( a + b ) \(⋮\)6

\(\Rightarrow\)a³ + b³ = ( a + b )³ - 3ab(a+b) 

Theo đề bài: ( a + b ) \(⋮\)6

\(\Rightarrow\)( a + b )³ \(⋮\)6

\(\Rightarrow3ab\left(a+b\right)⋮6\)

Mà a³ + b³ = ( a + b )³ - 3ab(a+b) 

\(\Rightarrow a^3+b^3⋮6\)

Vậy a³ + b³ chia hết cho 6 ( đpcm )