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\(\frac{4678\times4679+4680\times31+4648}{4680\times4679-4678\times4679}\)
\(\Leftrightarrow\frac{4679\times4678+4679\times31+4679}{4679\times\left(4680-4678\right)}\)
\(=\frac{4679\times\left(4678+31+1\right)}{4679\times2}\)
\(=\frac{4679\times4710}{4679\times2}\)
\(=\frac{4710}{2}\)
\(=2355\)
\(\frac{4678\cdot4679+4680\cdot31+4648}{4680\cdot4679-4678.4679}\)=\(\frac{4678\cdot4679+\left(4680-1\right)\cdot31+4648+31}{4680\cdot4679-4678.4679}\)
=\(\frac{4678\cdot4679+4679\cdot31+4679}{4680\cdot4679-4678.4679}\)=\(\frac{\left(4678+31+1\right)\cdot4679}{\left(4680-4678\right).4679}\)=\(\frac{4710}{2}\)=2355
a) \(\left(\frac{4}{3}-\frac{4}{6}\right)+\left(\frac{4}{6}-\frac{4}{9}\right)+\left(\frac{4}{9}-\frac{4}{10}\right)+\left(\frac{4}{12}-\frac{4}{15}\right)\)
\(=\frac{4}{15}-\frac{4}{3}=\frac{-16}{15}\)
C) bạn chỉ ần bỏ các số giống nhau thôi nhé
= 1
b)
Đặt biểu thức trên là A ta có:
A = \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{24}\)+ \(\frac{1}{48}\)+ \(\frac{1}{96}\)
A x 3 = \(1\)+ \(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)
A x 3 = \(1\)+ \(1\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{8}\)+ \(\frac{1}{8}\)- \(\frac{1}{16}\)+ \(\frac{1}{16}\)- \(\frac{1}{32}\)
A x 3 = 2 - \(\frac{1}{32}\)= \(\frac{63}{32}\)
A = \(\frac{63}{32}\): 3 = \(\frac{63}{96}\)
\(1\cdot\frac{1}{15}\cdot1\frac{1}{16}\cdot1\frac{1}{17}\cdot....\cdot1\frac{1}{2016}\cdot1\frac{1}{2017}\)
\(=\frac{1}{15}\cdot\frac{17}{16}\cdot\frac{18}{17}\cdot....\cdot\frac{2017}{2016}\cdot\frac{2018}{2017}\)
\(=\frac{1}{15}\cdot\frac{1}{16}\cdot2018\)
Dấu "." là dấu nhân nhé bn! phần còn lại bn làm tiếp nha
\(A=\frac{2019}{2}+\frac{2019}{6}+\frac{2019}{12}+....+\frac{2019}{2018.2019}\)
\(=\frac{2019}{1}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2018.2019}\right)\)
\(=\frac{2019}{1}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)
\(=\frac{2019}{1}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+....+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=\frac{2019}{1}.\left(1-\frac{1}{2019}\right)\)
\(=\frac{2019}{1}.\frac{2018}{2019}\)
\(=2018\)
\(A=\frac{2019}{2}+\frac{2019}{6}+\frac{2019}{12}+\frac{2019}{20}+\frac{2019}{30}+\frac{2019}{2018.2019}\)
\(A=\frac{2019}{1.2}+\frac{2019}{2.3}+\frac{2019}{3.4}+\frac{2019}{4.5}+\frac{2019}{5.6}+...+\frac{2019}{2018.2019}\)
\(A=2019.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)
\(A=2019.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(A=2019.\left(1-\frac{1}{2019}\right)\)\(=2019.\frac{2018}{2019}=2018\)
Vậy A = 2018
-Dấu " . " là dấu nhân.
ĐẶT BIỂU THỨC TRÊN LÀ M
TA CÓ \(2M=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{64}\)
\(\Rightarrow2M-M=1+\frac{1}{2}+\frac{1}{4}+..+\frac{1}{64}-\frac{1}{2}+\frac{1}{4}+..+\frac{1}{128}\)
\(\Rightarrow M=1+\frac{1}{28}\)
A= \(\frac{1}{2}\)+\(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
2A=2(\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\))
=1+\(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)
2A-A= (\(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)) -(\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\))
A=1-\(\frac{1}{128}\)
A=\(\frac{127}{128}\)
a) \(\frac{2}{11x16}+\frac{2}{16x21}+...+\frac{2}{61x66}\)
\(=\frac{2}{5}x\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\right)\)
\(=\frac{2}{5}x\left(\frac{1}{11}-\frac{1}{66}\right)\)
\(=\frac{2}{5}x\frac{5}{66}\)
\(=\frac{1}{33}\)
b) \(\frac{2}{5x7}+\frac{4}{7x11}+\frac{3}{11x14}+\frac{4}{14x18}+\frac{5}{18x23}+\frac{7}{23x30}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)
\(=\frac{1}{5}-\frac{1}{30}\)
\(=\frac{1}{6}\)
a, \(\frac{2}{11\times16}+\frac{2}{16\times21}+...+\frac{2}{61\times66}\)
\(=\frac{2}{5}\times\left(\frac{5}{11\times16}+...+\frac{5}{61\times66}\right)\)
\(=\frac{2}{5}\times\left(\frac{1}{11}-\frac{1}{16}+...+\frac{1}{61}-\frac{1}{66}\right)\)
\(=\frac{2}{5}\times\left(\frac{1}{11}-\frac{1}{66}\right)\)
\(=\frac{2}{5}\times\frac{5}{66}\)
\(=\frac{1}{33}\)
Vậy giá trị của biểu thức trên là : \(\frac{1}{33}\)
b,\(\frac{2}{5\times7}+\frac{4}{7\times11}+\frac{3}{11\times14}+\frac{4}{14\times18}+\frac{5}{18\times23}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}\)
\(=\frac{1}{5}-\frac{1}{23}\)
\(=\frac{18}{115}\)
Vậy giá trị của biểu thức trên là \(\frac{18}{115}\)
\(\frac{4678.4679+4680.31+4648}{4680.4679-4687.4679}\)
\(=\frac{4678.4679+4679.31+(31+4648)}{4679.\left(4680-4687\right)}\)
\(=\frac{4678.4679+4679.31+4679}{4679.\left(-7\right)}\)
\(=\frac{4679.\left(4678+31+1\right)}{4679.\left(-7\right)}\)
\(=\frac{4679.4710}{4679.\left(-7\right)}=\frac{4710}{-7}=\frac{-4710}{7}\)
lp 5 chưa hok đến số âm, nhưng bài này lại ra kp âm! nên b xem lại đề giúp mk nha!