1, Tính giá trị biểu thức :
\(a,A=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}.4\frac{1}{2}-2.2\frac{1}{3}\right):\frac{7}{4}\)
\(b,B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).............\left(1-\frac{1}{2017}\right).\left(1-\frac{1}{2018}\right)\)
Mọi người giúp mình giả toán nha !
Ta có:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)........\left(1-\frac{1}{2017}\right).\left(1-\frac{1}{2018}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{2016}{2017}.\frac{2017}{2018}\)
Đởn giản hết sẽ còn là:
\(\Rightarrow B=\frac{1}{2018}\)
Tính :
a) \(\text{A}=\left(1\times2\right)^{-1}+\left(2\times3\right)^{-1}+...+\left(2014\times2015\right)^{-1}\).
b) \(\text{B}=\frac{2018+\frac{2017}{2}+\frac{2016}{3}+\frac{2015}{4}+...+\frac{2}{2017}+\frac{1}{2018}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2018}+\frac{1}{2019}}\).
Thực hiện phép tính:
a, A= \(\frac{1}{6}.\left(-2\frac{3}{5}\right)+1\frac{2}{3}.\left(\frac{-13}{5}\right)\)
b, B=\(\frac{5}{7}:\left(\frac{1}{2}-\frac{1}{3}\right)+\frac{5}{7}:\left(\frac{1}{5}-\frac{1}{6}\right)\)
c, C= \(2^3+3.\left(\frac{2017}{2018}\right)^0.\left(\frac{1}{2}\right)^{-2}:4+[\left(-2\right)^2:\frac{1}{2}].\left(\frac{-1}{2}\right)^3\)
giúp vs, mik cần gấp
A= E387E4837
B = 883433
C = UỲUWFHQWURY48E3947
Câu 1. Tính hợp lý giá trị các biểu thức sau :
a. A = ( 689 - 31 ) - ( 269 - 131 )
b. B = \(\left(\frac{1}{2}+\frac{2016}{2017}+\frac{2017}{2018}+1\right)\times\left(\frac{2016}{2017}+\frac{2017}{2018}+\frac{3}{4}\right)-\left(\frac{1}{2}+\frac{2016}{2017}+\frac{2017}{2018}\right)\times\left(\frac{2016}{2017}+\frac{2017}{2018}+\frac{3}{4}+1\right)\)c. C = \(1-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\cdot\left(1-\frac{1}{2018}\right)\)
= (1/2).(2/3).(4/5).(5/6)......(2016/2017).(2017/2018)
=1.2.3.4.5......2016.2017/2.3.4.5.....2017.2018
=1/2018
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\cdot\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(=\frac{1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot2016\cdot2017}{2\cdot3\cdot4\cdot\cdot\cdot\cdot2017\cdot2018}\)
\(=\frac{1}{2018}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2016}{2017}.\frac{2017}{2018}\)
\(=\frac{1}{2018}\)
p/s: chúc bạn hok tốt
Cho A = \(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2017^2}-1\right)\cdot\left(\frac{1}{2018^2}-1\right)\) và B = \(-\frac{1}{2}\)
Hãy so sánh A và B
\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x......x\left(1-\frac{1}{2017}\right)x\left(1-\frac{1}{2018}\right)\)
=\(\frac{1}{2}x\frac{2}{3}x...x\frac{2017}{2018}\)
=\(\frac{1}{2018}\)
bạn trừ ra là đc
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{2017}\right)\cdot\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(=\frac{1\cdot2\cdot3\cdot....\cdot2016\cdot2017}{2\cdot3\cdot4\cdot....\cdot2017\cdot2018}\)
\(=\frac{1}{2018}\)
dấu lớn hơn vì 23/12 lớn hơn 2008/2009
\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x......x\left(1-\frac{1}{2017}\right)x\left(1-\frac{1}{2018}\right)\)
\(=\left(\frac{1}{2}\right).\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)....\left(\frac{2017}{2018}\right)\)
\(=\frac{1.2.3....2017}{2.3.4...2018}\)
\(=\frac{1}{2018}\)
\(=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2017}{2018}\)
\(=\frac{1}{2018}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}.\frac{2}{3}...\frac{2016}{2017}.\frac{2017}{2018}\)
\(=\frac{1.2.3.4...2016.2017}{2.3.4...2017.2018}=\frac{1}{2018}\)
Cho A= \(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right) \cdot \left(\frac{1}{4^2}-1\right)....\left(\frac{1}{2017^2}-1\right)\left(\frac{1}{2018^2}-1\right)\) và B = \(\frac{-1}{2}\)
