Tính
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(\frac{1}{3}+\frac{1}{5}+\frac{1}{35}+...+\frac{1}{9999}\)
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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 )
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)