CMR:\(\frac{7}{3}+\frac{11}{3^2}+\frac{15}{3^3}+...+\frac{2019}{3^{504}}< \frac{9}{2}\)
Cho A=\(\frac{7}{3}+\frac{11}{3^2}+\frac{15}{3^3}+...\frac{2019}{3^{504}}\)
Chứng minh rằng A<9/2
Cho A=\(\frac{7}{3}+\frac{11}{3^2}+\frac{15}{3^3}+...+\frac{2019}{3^{504}}\)
Chứng tỏ A<\(\frac{9}{2}\)
Cho A= \(\frac{7}{3}\)+ \(\frac{11}{3^2}\)+ \(\frac{15}{3^3}\)+ .......+ \(\frac{2019}{3^{504}}\). Chứng minh rằng A< \(\frac{9}{2}\)
Cho đa thức : f(x)=x(x^19-x^5-x^2018) và g(x)= x^2019-x^2020+9+(x^4+x^2+2)
1)Tính k(x)=f(x)+g(x)
2)Tính giá trị của k(x) tại x bằng \(\left(2-\frac{5}{3}+\frac{7}{6}-\frac{9}{10}+\frac{11}{15}-\frac{13}{21}+\frac{15}{28}-\frac{17}{36}+\frac{19}{45}\right)\cdot\frac{5}{6}\)
3) CMR k(x) không nhận giá trị 2019 với mọi giá trị nguyên x
B=\(\frac{1}{3}-\frac{3}{4}-\left(-0,6\right)+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
C=\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{13}{15}+\frac{11}{15}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+0,6-\frac{1}{3}\)
D=\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}.....-\frac{1}{3.2}-\frac{1}{2.1}\)
\(B=\frac{1}{3}-\frac{3}{4}+0,6+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow B=\frac{3}{15}-\frac{48}{64}+\frac{9}{15}+\frac{1}{64}-\frac{8}{36}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow B=\frac{3}{15}+\frac{9}{15}+\frac{1}{15}+\left(-\frac{48}{64}+\frac{1}{64}\right)+\left(-\frac{8}{36}-\frac{1}{36}\right)\)
\(\Rightarrow B=\frac{13}{15}-\frac{47}{64}-\frac{1}{4}\)
\(\Rightarrow B=-\frac{113}{960}\)
\(C=0\)
\(D=\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow D=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+...-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+1\)
\(\Rightarrow D=1\)
D= \(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}......-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{97.98}+\frac{1}{98.99}\right)\)
=\(\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-.....-\frac{1}{98}-\frac{1}{99}\right)\)
=\(\frac{1}{99}-\left[1-(\frac{1}{2}-\frac{1}{2}+......+\frac{1}{98}-\frac{1}{99})\right]\)
=\(\frac{1}{99}-\left(1-0-0-.....-0-\frac{1}{99}\right)\)
=\(\frac{1}{99}-1-\frac{1}{99}\)
=1
\(A=\frac{1}{3}-\frac{3}{4}-\left(-\frac{3}{5}\right)+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)\(\frac{1}{15}\)
\(B=\frac{1}{5}-\frac{3}{7}+\frac{5}{9}-\frac{2}{11}+\frac{7}{13}-\frac{9}{16}-\frac{7}{13}+\frac{2}{11}-\frac{5}{9}+\frac{3}{7}-\frac{1}{5}\)
Cho A = \(\frac{2}{3^2}\)+ \(\frac{2}{5^2}\)+\(\frac{2}{7^2}\)+...+\(\frac{2}{2019^2}\). CMR : A < \(\frac{504}{1009}\)
ta có:
2/3^2+2/5^2+...+2/2019^2 < 2/(3.5)+2/(5.7)+...+2/(2019.2021)
=> A < 1/3-1/5+...+1/2019-1/2021
=> A < 1/3-1/2021
=> A <2018/6063
=> A <2520/6063 - 520/6063 (1)
Vì 2520/6063<504/1009=>2520/6063 - 502/6063 <504/1009 (2)
Từ (1) và (2) => A< 504/1009
Sabofans, cảm ơn bạn nhưng bạn làm hơi khó hiểu 1 chút ><
\(\left(8-\frac{9}{4}+\frac{2}{7}\right)-\left(-6-\frac{3}{7}+\frac{5}{4}\right)-\left(3+\frac{2}{4}-\frac{9}{7}\right)\)\(\frac{9}{7}\))
\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{13}{15}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)
\(\frac{1}{2014}-\frac{1}{2014.2013}-\frac{1}{2013.2012}-\frac{1}{2012.2011}-...-\frac{1}{3.2}-\frac{1}{2.1}\)